5 Multiple Integration 多重积分

Chapter Outline 本章概要

• 5.1 Double Integrals over Rectangular Regions
• 5.2 Double Integrals over General Regions
• 5.3 Double Integrals in Polar Coordinates
• 5.4 Triple Integrals
• 5.5 Triple Integrals in Cylindrical and Spherical Coordinates
• 5.6 Calculating Centers of Mass and Moments of Inertia
• 5.7 Change of Variables in Multiple Integrals

• 5.1 矩形区域上的二重积分
• 5.2 一般区域上的二重积分
• 5.3 极坐标下的二重积分
• 5.4 三重积分
• 5.5 柱坐标与球坐标下的三重积分
• 5.6 质心与转动惯量的计算
• 5.7 多重积分中的变量代换

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5.1 Double Integrals over Rectangular Regions 5.1 矩形区域上的二重积分

• 5.1.1 Recognize when a function of two variables is integrable over a rectangular region.
• 5.1.2 Recognize and use some of the properties of double integrals.
• 5.1.3 Evaluate a double integral over a rectangular region by writing it as an iterated integral.
• 5.1.4 Use a double integral to calculate the area of a region, volume under a surface, or average value of a function over a plane region.

• 5.1.1 判别二元函数何时在矩形区域上可积。
• 5.1.2 认识并运用二重积分的若干性质。
• 5.1.3 把矩形区域上的二重积分写成累次积分来求值。
• 5.1.4 用二重积分计算区域的面积、曲面下的体积，或函数在平面区域上的平均值。

In this section we investigate double integrals and show how we can use them to find the volume of a solid over a rectangular region in the $xy$-plane. Many of the properties of double integrals are similar to those we have already discussed for single integrals.

Volumes and Double Integrals 体积与二重积分

We begin by considering the space above a rectangular region *R*. Consider a continuous function $f\left( {x,y} \right) \geq 0$ of two variables defined on the closed rectangle *R*:

$$R = \lbrack a,b\rbrack\ \times \ \lbrack c,d\rbrack = \left\{ (x,y) \in \mathbb{R}^{2}\left| \left. a \leq x \leq b,c \leq y \leq d \right\} \right. \right.$$

Here $\lbrack a,b\rbrack\ \times \ \lbrack c,d\rbrack$ denotes the Cartesian product of the two closed intervals $\lbrack a,b\rbrack$ and $\lbrack c,d\rbrack.$ It consists of rectangular pairs $(x,y)$ such that $a \leq x \leq b$ and $c \leq y \leq d.$ The graph of $f$ represents a surface above the $xy$-plane with equation $z = f\left( {x,y} \right)$ where $z$ is the height of the surface at the point $(x,y).$ Let $S$ be the solid that lies above $R$ and under the graph of $f$ (Figure 5.2). The base of the solid is the rectangle $R$ in the $xy$-plane. We want to find the volume $V$ of the solid $S.$

We divide the region $R$ into small rectangles $R_{ij},$ each with area $\text{Δ}A$ and with sides $\text{Δ}x$ and $\text{Δ}y$ (Figure 5.3). We do this by dividing the interval $\lbrack a,b\rbrack$ into $m$ subintervals and dividing the interval $\lbrack c,d\rbrack$ into $n$ subintervals. Hence $\text{Δ}x = \frac{b - a}{m},$ $\text{Δ}y = \frac{d - c}{n},$ and $\text{Δ}A = \text{Δ}x\text{Δ}y.$

The volume of a thin rectangular box above $R_{ij}$ is $f(x_{ij}^{*},y_{ij}^{*})\text{Δ}A,$ where $(x_{ij}^{*},y_{ij}^{*})$ is an arbitrary sample point in each $R_{ij}$ as shown in the following figure.

Using the same idea for all the subrectangles, we obtain an approximate volume of the solid $S$ as $V \approx {\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f(x_{ij}^{*},y_{ij}^{*})\text{Δ}A}}}.$ This sum is known as a double Riemann sum and can be used to approximate the value of the volume of the solid. Here the double sum means that for each subrectangle we evaluate the function at the chosen point, multiply by the area of each rectangle, and then add all the results.

As we have seen in the single-variable case, we obtain a better approximation to the actual volume if *m* and *n* become larger.

$$V = \underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f(x_{ij}^{*},y_{ij}^{*})\text{Δ}}}}A\ \text{or}\ V = \underset{\text{Δ}x,\text{Δ}y\rightarrow 0}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f(x_{ij}^{*},y_{ij}^{*})\text{Δ}A.}}}$$

Note that the sum approaches a limit in either case and the limit is the volume of the solid with the base *R*. Now we are ready to define the double integral.

The double integral of the function $f\left( {x,y} \right)$ over the rectangular region $R$ in the $xy$-plane is defined as

$${\iint\limits_{R}{f(x,y)dA}} = \underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f(x_{i}^{*},y_{j}^{*})\text{Δ}A}}}.$$ (5.1)

If $f\left( {x,y} \right) \geq 0,$ then the volume *V* of the solid *S*, which lies above $R$ in the $xy$-plane and under the graph of *f*, is the double integral of the function $f\left( {x,y} \right)$ over the rectangle $R.$ If the function is ever negative, then the double integral can be considered a “signed” volume in a manner similar to the way we defined net signed area in The Definite Integral.

Setting up a Double Integral and Approximating It by Double Sums 建立二重积分并用二重和作近似

Consider the function $z = f(x,y) = 3x^{2} - y$ over the rectangular region $R = \lbrack 0,2\rbrack\ \times \ \lbrack 0,2\rbrack$ (Figure 5.5).

1. Set up a double integral for finding the value of the signed volume of the solid *S* that lies above $R$ and “under” the graph of $f.$

2. Divide *R* into four squares with $m = n = 2,$ and choose the sample point as the upper right corner point of each square $(1,1),(2,1),(1,2),$ and $(2,2)$ (Figure 5.6) to approximate the signed volume of the solid *S* that lies above $R$ and “under” the graph of $f.$

3. Divide *R* into four squares with $m = n = 2,$ and choose the sample point as the midpoint of each square: $(1\text{/}2,1\text{/}2),(3\text{/}2,1\text{/}2),(1\text{/}2,3\text{/}2),\text{and}\ (3\text{/}2,3\text{/}2)$ to approximate the signed volume.

Solution 解

1. As we can see, the function $z = f(x,y) = 3x^{2} - y^{}$ is both above and below the plane. To find the signed volume of *S*, we need to divide the region *R* into small rectangles $R_{ij},$ each with area $\text{Δ}A$ and with sides $\text{Δ}x$ and $\text{Δ}y,$ and choose $(x_{ij}^{*},y_{ij}^{*})$ as sample points in each $R_{ij}.$ Hence, a double integral is set up as

$$V = {\iint\limits_{R}{\left( {3x^{2} - y} \right)dA}} = \underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{\left\lbrack {3\left( x_{ij}^{*} \right)^{2} - y_{ij}^{*}} \right\rbrack\text{Δ}A}}}.$$

2. Approximating the signed volume using a Riemann sum with $m = n = 2$ we have $\text{Δ}A = \text{Δ}x\text{Δ}y = 1\ \times \ 1 = 1.$ Also, the sample points are (1, 1), (2, 1), (1, 2), and (2, 2) as shown in the following figure.

Hence,

$$\begin{matrix} V & {= \sum\limits_{i = 1}^{2}\sum\limits_{j = 1}^{2}{f\left( x_{ij}^{*},y_{ij}^{*} \right)\text{Δ}A}} \\ & {= \sum\limits_{i = 1}^{2}\left( f\left( x_{i1}^{*},y_{i1}^{*} \right) + \right.f\left( x_{i2}^{*},y_{i2}^{*} \right))\text{Δ}A} \\ & {= f\left( x_{11}^{*},y_{11}^{*} \right)\text{Δ}A + f\left( x_{21}^{*},y_{21}^{*} \right)\text{Δ}A + f\left( x_{12}^{*},y_{12}^{*} \right)\text{Δ}A + f\left( x_{22}^{*},y_{22}^{*} \right)\text{Δ}A} \\ & {= f(1,1)(1) + f(2,1)(1) + f(1,2)(1) + f(2,2)(1)} \\ & {= \left( {3 - 1} \right)(1) + (12 - 1)(1) + (3 - 2)(1) + (12 - 2)(1)} \\ & {= 2~ + ~11~ + ~1~ + ~10~ = ~24} \end{matrix}$$

3. Approximating the signed volume using a Riemann sum with $m = n = 2,$ we have $\text{Δ}A = \text{Δ}x\text{Δ}y = 1\ \times \ 1 = 1.$ In this case the sample points are (1/2, 1/2), (3/2, 1/2), (1/2, 3/2),

and (3/2, 3/2).

Hence

$$\begin{matrix} V & {= \sum\limits_{i = 1}^{2}\sum\limits_{j = 1}^{2}{f\left( x_{ij}^{*},y_{ij}^{*} \right)\text{Δ}A}} \\ & {= f\left( x_{11}^{*},y_{11}^{*} \right)\text{Δ}A + f\left( x_{21}^{*},y_{21}^{*} \right)\text{Δ}A + f\left( x_{12}^{*},y_{12}^{*} \right)\text{Δ}A + f\left( x_{22}^{*},y_{22}^{*} \right)\text{Δ}A} \\ & {= f\left( \frac{1}{2},\frac{1}{2} \right)(1) + f\left( \frac{3}{2},\frac{1}{2} \right)(1) + f\left( \frac{1}{2},\frac{3}{2} \right)(1) + f\left( \frac{3}{2},\frac{3}{2} \right)(1)} \\ & {= \left( \frac{3}{4} - \frac{1}{2} \right)(1) + \left( \frac{27}{4} - \frac{1}{2} \right)(1) + \left( \frac{3}{4} - \frac{3}{2} \right)(1) + \left( \frac{27}{4} - \frac{3}{2} \right)(1)} \\ & {= \frac{1}{4} + \frac{25}{4} + \left( - \frac{3}{4} \right) + \frac{21}{4} = \frac{44}{4} = 11} \end{matrix}$$

Analysis 分析

Notice that the approximate answers differ due to the choices of the sample points. In either case, we are introducing some error because we are using only a few sample points. Thus, we need to investigate how we can achieve an accurate answer.

Use the same function $z = f(x,y) = 3x^{2} - y^{}$ over the rectangular region $R = \lbrack 0,2\rbrack\ \times \ \lbrack 0,2\rbrack.$

Divide *R* into the same four squares with $m = n = 2,$ and choose the sample points as the upper left corner point of each square $(0,1),(1,1),(0,2),$ and $(1,2)$ (Figure 5.6) to approximate the signed volume of the solid *S* that lies above $R$ and “under” the graph of $f.$

Note that we developed the concept of double integral using a rectangular region *R*. This concept can be extended to any general region. However, when a region is not rectangular, the subrectangles may not all fit perfectly into *R*, particularly if the base area is curved. We examine this situation in more detail in the next section, where we study regions that are not always rectangular and subrectangles may not fit perfectly in the region *R*. Also, the heights may not be exact if the surface $z = f\left( {x,y} \right)$ is curved. However, the errors on the sides and the height where the pieces may not fit perfectly within the solid *S* approach 0 as *m* and *n* approach infinity. Also, the double integral of the function $z = f\left( {x,y} \right)$ exists provided that the function $f$ is not too discontinuous. If the function is bounded and continuous over *R* except on a finite number of smooth curves, then the double integral exists and we say that $f$ is integrable over *R*.

Since $\text{Δ}A = \text{Δ}x\text{Δ}y = \text{Δ}y\text{Δ}x,$ we can express $dA$ as $dx\ dy$ or $dy\ dx.$ This means that, when we are using rectangular coordinates, the double integral over a region $R$ denoted by ${\iint\limits_{R}{f(x,y)d}}A$ can be written as ${\iint\limits_{R}{f(x,y)d}}x\ dy$ or ${\iint\limits_{R}{f(x,y)d}}y\ dx.$

Now let’s list some of the properties that can be helpful to compute double integrals.

Properties of Double Integrals 二重积分的性质

The properties of double integrals are very helpful when computing them or otherwise working with them. We list here six properties of double integrals. Properties 1 and 2 are referred to as the linearity of the integral, property 3 is the additivity of the integral, property 4 is the monotonicity of the integral, and property 5 is used to find the bounds of the integral. Property 6 is used if $f(x,y)$ is a product of two functions $g(x)$ and $h(y).$

Properties of Double Integrals 二重积分的性质

Assume that the functions $f(x,y)$ and $g(x,y)$ are integrable over the rectangular region *R*; *S* and *T* are subregions of *R*; and assume that *m* and *M* are real numbers.

1. The sum $f(x,y) + g(x,y)$ is integrable and

$${\iint\limits_{R}{\left\lbrack {f(x,y) + g(x,y)} \right\rbrack dA =}}{\iint\limits_{R}{f(x,y)dA +}}{\iint\limits_{R}{g(x,y)dA.}}$$

2. If *c* is a constant, then $cf(x,y)$ is integrable and

$${\iint\limits_{R}{cf(x,y)dA = c}}{\iint\limits_{R}{f(x,y)dA.}}$$

3. If $R = S \cup T$ and $S \cap T = \varnothing$ except an overlap on the boundaries, then

$${\iint\limits_{R}{f(x,y)dA =}}{\iint\limits_{S}{f(x,y)dA + {\iint\limits_{T}{f(x,y)dA}}.}}$$

4. If $f(x,y) \geq g(x,y)$ for $(x,y)$ in $R,$ then

$${\iint\limits_{R}{f(x,y)dA \geq}}{\iint\limits_{R}{g(x,y)dA.}}$$

5. If $m \leq f(x,y) \leq M,$ then

$$m\ \times \ A(R) \leq {\iint\limits_{R}{f(x,y)dA \leq M\ \times \ A(R)}}.$$

6. In the case where $f(x,y)$ can be factored as a product of a function $g(x)$ of $x$ only and a function $h(y)$ of $y$ only, then over the region $R = \left\{ \left( {x,y} \right) \middle| a \leq x \leq b,c \leq y \leq d \right\},$ the double integral can be written as

$${\iint\limits_{R}{f(x,y)dA}} = \left( {\int_{a}^{b}{g(x)dx}} \right)\left( {\int_{c}^{d}{h(y)dy}} \right).$$

These properties are used in the evaluation of double integrals, as we will see later. We will become skilled in using these properties once we become familiar with the computational tools of double integrals. So let’s get to that now.

Iterated Integrals 累次积分

So far, we have seen how to set up a double integral and how to obtain an approximate value for it. We can also imagine that evaluating double integrals by using the definition can be a very lengthy process if we choose larger values for $m$ and $n.$ Therefore, we need a practical and convenient technique for computing double integrals. In other words, we need to learn how to compute double integrals without employing the definition that uses limits and double sums.

The basic idea is that the evaluation becomes easier if we can break a double integral into single integrals by integrating first with respect to one variable and then with respect to the other. The key tool we need is called an iterated integral.

Assume $a,b,c,$ and $d$ are real numbers. We define an iterated integral for a function $f(x,y)$ over the rectangular region $R$ $= \lbrack a,b\rbrack\ \times \ \lbrack c,d\rbrack$ as

1.

$${\int\limits_{a}^{b}\ {\int\limits_{c}^{d}{f(x,y)dy\ dx}}} = {\int\limits_{a}^{b}\left\lbrack {\int\limits_{c}^{d}{f(x,y)dy}} \right\rbrack}dx$$ (5.2)

2.

$${\int\limits_{c}^{d}\ {\int\limits_{a}^{b}{f(x,y)dx\ dy}}} = {\int\limits_{c}^{d}\left\lbrack {\int\limits_{a}^{b}{f(x,y)dx}} \right\rbrack}dy.$$ (5.3)

The notation ${\int\limits_{a}^{b}\left\lbrack {\int\limits_{c}^{d}{f(x,y)dy}} \right\rbrack}dx$ means that we integrate $f(x,y)$ with respect to *y* while holding *x* constant. Similarly, the notation ${\int\limits_{c}^{d}\left\lbrack {\int\limits_{a}^{b}{f(x,y)dx}} \right\rbrack}dy$ means that we integrate $f(x,y)$ with respect to *x* while holding *y* constant. The fact that double integrals can be split into iterated integrals is expressed in Fubini’s theorem. Think of this theorem as an essential tool for evaluating double integrals.

Fubini’s Theorem 富比尼定理

Suppose that $f(x,y)$ is a function of two variables that is continuous over a rectangular region $R = \left\{ {\left( {x,y} \right) \in \mathbb{R}^{2}\left| {a \leq x \leq b,\text{c} \leq \text{y} \leq \text{d}} \right.} \right\}.$ Then we see from Figure 5.7 that the double integral of $f$ over the region equals an iterated integral,

$${\iint\limits_{R}{f(x,y)dA =}}{\iint\limits_{R}{f(x,y)dx\ dy}} = {\int\limits_{a}^{b}\ {\int\limits_{c}^{d}{f(x,y)dy\ dx}}} = {\int\limits_{c}^{d}\ {\int\limits_{a}^{b}{f(x,y)dx\ dy}}}.$$

More generally, Fubini’s theorem is true if $f$ is bounded on $R$ and $f$ is discontinuous only on a finite number of continuous curves. In other words, $f$ has to be integrable over $R.$

Using Fubini’s Theorem 富比尼定理的应用

Use Fubini’s theorem to compute the double integral $\iint\limits_{R}{f(x,y)dA}$ where $f(x,y) = x$ and $R = \lbrack 0,2\rbrack\ \times \ \lbrack 0,1\rbrack.$

Solution 解

Fubini’s theorem offers an easier way to evaluate the double integral by the use of an iterated integral. Note how the boundary values of the region *R* become the upper and lower limits of integration.

$$\begin{array}{cl} {\iint\limits_{R}{f(x,y)dA}} & {= {\iint\limits_{R}{f(x,y)dx\ dy}}} \\ & {= {\int_{y = 0}^{y = 1}{\int_{x = 0}^{x = 2}{x\ dx\ dy}}}} \\ & {= {\int_{y = 0}^{y = 1}\left\lbrack \left. \frac{x^{2}}{2} \right|_{x = 0}^{x = 2} \right\rbrack}dy} \\ & {= {\int_{y = 0}^{y = 1}{2dy = \left. {2y} \right|_{y = 0}^{y = 1} = 2}}.} \end{array}$$

The double integration in this example is simple enough to use Fubini’s theorem directly, allowing us to convert a double integral into an iterated integral. Consequently, we are now ready to convert all double integrals to iterated integrals and demonstrate how the properties listed earlier can help us evaluate double integrals when the function $f(x,y)$ is more complex. Note that the order of integration can be changed (see Example 5.7).

Illustrating Properties i and ii 性质 i 与性质 ii 的示例

Evaluate the double integral ${\iint\limits_{R}\left( {xy - 3xy^{2}} \right)}dA$ where $R = \left\{ \left( {x,y} \right) \middle| 0 \leq x \leq 2,1 \leq y \leq 2 \right\}.$

Solution 解

This function has two pieces: one piece is $xy$ and the other is $3xy^{2}.$ Also, the second piece has a constant $3.$ Notice how we use properties i and ii to help evaluate the double integral.

$$\begin{array}{lccc} & & & \\ & & & \\ & & & \\ & & & \\ {\quad{\iint\limits_{R}\left( {xy - 3xy^{2}} \right)}dA} & & & \\ {= {\iint\limits_{R}{xy}}\ dA + {\iint\limits_{R}\left( {-3xy^{2}} \right)}dA} & & & \text{Property i: Integral of a sum is the sum of the integrals.} \\ {= {\int_{y = 1}^{y = 2}{\int_{x = 0}^{x = 2}{xy\ dx\ dy}}} - {\int_{y = 1}^{y = 2}{\int_{x = 0}^{x = 2}{3xy^{2}dx\ dy}}}} & & & \text{Convert double integrals to iterated integrals.} \\ {= {\int_{y = 1}^{y = 2}{\left. \left( {\frac{x^{2}}{2}y} \right) \right|_{x = 0}^{x = 2}dy}} - 3{\int_{y = 1}^{y = 2}{\left. \left( {\frac{x^{2}}{2}y^{2}} \right) \right|_{x = 0}^{x = 2}dy}}} & & & {\text{Integrate with respect to}\ x,\text{holding}\ y\ \text{constant.}} \\ {= {\int_{y = 1}^{y = 2}{2y\ dy}} - {\int_{y = 1}^{y = 2}{6y^{2}dy}}} & & & \text{Property ii: Placing the constant before the integral.} \\ {= {2\int_{1}^{2}{y\ dy}} - 6{\int_{1}^{2}{y^{2}dy}}} & & & {\text{Integrate with respect to}\ y.} \\ {= \left. {2\frac{y^{2}}{2}} \right|_{1}^{2} - \left. {6\frac{y^{3}}{3}} \right|_{1}^{2}} & & & \\ {= \left. y^{2} \right|_{1}^{2} - \left. {2y^{3}} \right|_{1}^{2}} & & & \\ {= (4 - 1) - 2(8 - 1)} & & & \\ {= 3 - 2(7) = 3 - 14 = -11.} & & & \end{array}$$

Illustrating Property v. 性质 v 的示例

Over the region $R = \left\{ \left( {x,y} \right) \middle| 1 \leq x \leq 3,1 \leq y \leq 2 \right\},$ we have $2 \leq x^{2} + y^{2} \leq 13.$ Find a lower and an upper bound for the integral ${\iint\limits_{R}\left( {x^{2} + y^{2}} \right)}dA.$

Solution 解

For a lower bound, integrate the constant function 2 over the region $R.$ For an upper bound, integrate the constant function 13 over the region $R.$

$$\begin{array}{cll} {\int_{1}^{2}{\int_{1}^{3}{2dx\ dy}}} & = & {{\int_{1}^{2}\left\lbrack \left. {2x} \right|_{1}^{3} \right\rbrack}dy = {\int_{1}^{2}{2(2)dy}} = \left. {4y} \right|_{1}^{2} = 4(2 - 1) = 4} \\ {\int_{1}^{2}{\int_{1}^{3}{13dx\ dy}}} & = & {{\int_{1}^{2}\left\lbrack \left. {13x} \right|_{1}^{3} \right\rbrack}dy = {\int_{1}^{2}{13(2)dy}} = \left. {26y} \right|_{1}^{2} = 26(2 - 1) = 26.} \end{array}$$

Hence, we obtain $4 \leq {\iint\limits_{R}\left( {x^{2} + y^{2}} \right)}dA \leq 26.$

Illustrating Property vi 性质 vi 的示例

Evaluate the integral $\iint\limits_{R}{e^{y}\text{cos}\ x\ dA}$ over the region $R = \left\{ \left( {x,y} \right) \middle| 0 \leq x \leq \frac{\pi}{2},0 \leq y \leq 1 \right\}.$

Solution 解

This is a great example for property vi because the function $f(x,y)$ is clearly the product of two single-variable functions $e^{y}$ and $\text{cos}\ x.$ Thus we can split the integral into two parts and then integrate each one as a single-variable integration problem.

$$\begin{array}{cl} {\iint\limits_{R}{e^{y}\text{cos}\ x\ dA}} & {= {\int_{0}^{1}{\int_{0}^{\pi\text{/}2}{e^{y}\text{cos}\ x\ dx\ dy}}}} \\ & {= \left( {\int_{0}^{1}{e^{y}dy}} \right)\left( {\int_{0}^{\pi\text{/}2}{\text{cos}\ x\ dx}} \right)} \\ & {= \left( \left. e^{y} \right|_{0}^{1} \right)\left( \left. {\text{sin}\ x} \right|_{0}^{\pi\text{/}2} \right)} \\ & {= e - 1.} \end{array}$$

1. Use the properties of the double integral and Fubini’s theorem to evaluate the integral

$${\int_{0}^{1}{\int_{-1}^{3}{\left( {3 - x + 4y} \right)dy\ dx}}}.$$

2. Show that $0 \leq {\iint\limits_{R}{\text{sin}\ \pi x\ \text{cos}\ \pi y\ dA \leq \frac{1}{32}}}$ where $R = {\left\lbrack {0,\frac{1}{4}} \right\rbrack \times}\left\lbrack {\frac{1}{4},\frac{1}{2}} \right\rbrack.$

As we mentioned before, when we are using rectangular coordinates, the double integral over a region $R$ denoted by $\iint\limits_{R}{f(x,y)dA}$ can be written as ${\iint\limits_{R}{f(x,y)d}}x\ dy$ or ${\iint\limits_{R}{f(x,y)d}}y\ dx.$ The next example shows that the results are the same regardless of which order of integration we choose.

Evaluating an Iterated Integral in Two Ways 用两种方法计算累次积分

Let’s return to the function $f(x,y) = 3x^{2} - y$ from Example 5.1, this time over the rectangular region $R = \lbrack 0,2\rbrack\ \times \ \lbrack 0,3\rbrack.$ Use Fubini’s theorem to evaluate $\iint\limits_{R}{f(x,y)dA}$ in two different ways:

1. First integrate with respect to *y* and then with respect to *x*;

2. First integrate with respect to *x* and then with respect to *y*.

Solution 解

Figure 5.7 shows how the calculation works in two different ways.

1. First integrate with respect to *y* and then integrate with respect to *x*:

$$\begin{array}{cl} {\iint\limits_{R}{f(x,y)dA}} & {= {\int_{x = 0}^{x = 2}{\int_{y = 0}^{y = 3}{(3x^{2} - y)dy\ dx}}}} \\ & {= {\int_{x = 0}^{x = 2}\left( {\int\limits_{y = 0}^{y = 3}{(3x^{2} - y)dy}} \right)}dx = {\int_{x = 0}^{x = 2}{\left\lbrack \left. {3x^{2}y - \frac{y^{2}}{2}} \right|_{y = 0}^{y = 3} \right\rbrack dx}}} \\ & {= {\int_{x = 0}^{x = 2}\left( {9x^{2} - \frac{9}{2}} \right)}dx = \left. {3x^{3} - \frac{9}{2}x} \right|_{x = 0}^{x = 2} = 15.} \end{array}$$

2. First integrate with respect to *x* and then integrate with respect to *y*:

$$\begin{array}{cl} {\iint\limits_{R}{f(x,y)dA}} & {= {\int_{y = 0}^{y = 3}{\int_{x = 0}^{x = 2}{(3x^{2} - y)dx\ dy}}}} \\ & {= {\int_{y = 0}^{y = 3}\left( {\int_{x = 0}^{x = 2}{(3x^{2} - y)dx}} \right)}dy = {\int_{y = 0}^{y = 3}{\left\lbrack \left. {x^{3} - xy} \right|_{x = 0}^{x = 2} \right\rbrack dy}}} \\ & {= {\int_{y = 0}^{y = 3}{\left( {8 - 2y} \right)dy}} = \left. {8y - y^{2}} \right|_{y = 0}^{y = 3} = 15.} \end{array}$$

Analysis 分析

With either order of integration, the double integral gives us an answer of 15. We might wish to interpret this answer as a volume in cubic units of the solid $S$ below the function $f(x,y) = 3x^{2} - y$ over the region $R = \lbrack 0,2\rbrack\ \times \ \lbrack 0,3\rbrack.$ However, remember that the interpretation of a double integral as a (non-signed) volume works only when the integrand $f$ is a nonnegative function over the base region $R.$

Evaluate ${\int_{y = -3}^{y = 2}{\int_{x = 3}^{x = 5}{\left( {2 - 3x^{2} + y^{2}} \right)dx\ dy}}}.$

In the next example we see that it can actually be beneficial to switch the order of integration to make the computation easier. We will come back to this idea several times in this chapter.

Switching the Order of Integration 交换积分次序

Consider the double integral $\iint\limits_{R}{x\ \text{sin}(xy)dA}$ over the region $R = \left\{ \left( {x,y} \right) \middle| 0 \leq x \leq \pi,1 \leq y \leq 2 \right\}$ (Figure 5.8).

1. Express the double integral in two different ways.

2. Analyze whether evaluating the double integral in one way is easier than the other and why.

3. Evaluate the integral.

Solution 解

1. We can express $\iint\limits_{R}{x\ \text{sin}(xy)dA}$ in the following two ways: first by integrating with respect to $y$ and then with respect to $x;$ second by integrating with respect to $x$ and then with respect to $y.$

$$\begin{array}{lccc} & & & \\ & & & \\ & & & \\ & & & \\ {\quad{\iint\limits_{R}{x\ \text{sin}(xy)dA}}} & & & \\ {= {\int\limits_{x = 0}^{x = \pi}\ {\int\limits_{y = 1}^{y = 2}{x\ \text{sin}(xy)dy\ dx}}}} & & & {\text{Integrate first with respect to}\ y.} \\ {= {\int\limits_{y = 1}^{y = 2}\ {\int\limits_{x = 0}^{x = \pi}{x\ \text{sin}(xy)dx\ dy}}}} & & & {\text{Integrate first with respect to}\ x.} \end{array}$$

2. If we want to integrate with respect to *y* first and then integrate with respect to $x,$ we see that we can use the substitution $u = xy,$ which gives $du = x\ dy.$ Hence the inner integral is simply ${\int{\text{sin}\ u}}\ du$ and we can change the limits to be functions of *x*,

$${\iint\limits_{R}{x\ \text{sin}(xy)dA}} = {\int\limits_{x = 0}^{x = \pi}\ {\int\limits_{y = 1}^{y = 2}{x\ \text{sin}(xy)dy\ dx}}} = {\int\limits_{x = 0}^{x = \pi}\left\lbrack {\int\limits_{u = x}^{u = 2x}{\text{sin}(u)du}} \right\rbrack}dx.$$

However, integrating with respect to $x$ first and then integrating with respect to $y$ requires integration by parts for the inner integral, with $u = x$ and $dv = \text{sin}(xy)dx.$

Then $du = dx$ and $v = - \frac{\text{cos}(xy)}{y},$ so

$${\iint\limits_{R}{x\ \text{sin}(xy)dA}} = {\int\limits_{y = 1}^{y = 2}\ {\int\limits_{x = 0}^{x = \pi}{x\ \text{sin}(xy)dx\ dy}}} = {\int\limits_{y = 1}^{y = 2}\left\lbrack {- \left. \frac{x\ \text{cos}(xy)}{y} \right|_{x = 0}^{x = \pi} + \frac{1}{y}{\int\limits_{x = 0}^{x = \pi}{\text{cos}(xy)dx}}} \right\rbrack}dy.$$

Since the evaluation is getting complicated, we will only do the computation that is easier to do, which is clearly the first method.

3. Evaluate the double integral using the easier way.

$$\begin{array}{cl} {\iint\limits_{R}{x\ \text{sin}(xy)dA}} & {= {\int\limits_{x = 0}^{x = \pi}\ {\int\limits_{y = 1}^{y = 2}{x\ \text{sin}(xy)dy\ dx}}}} \\ & {= {\int\limits_{x = 0}^{x = \pi}\left\lbrack {\int\limits_{u = x}^{u = 2x}{\text{sin}(u)du}} \right\rbrack}dx = {\int\limits_{x = 0}^{x = \pi}\left\lbrack \left. {\text{−}\text{cos}\ u} \right|_{u = x}^{u = 2x} \right\rbrack}dx = {\int\limits_{x = 0}^{x = \pi}{\left( {\text{−}\text{cos}\ 2x + \text{cos}\ x} \right)dx}}} \\ & {= \left. {- \frac{1}{2}\text{sin}\ 2x + \text{sin}\ x} \right|_{x = 0}^{x = \pi} = 0.} \end{array}$$

Evaluate the integral $\iint\limits_{R}{xe^{xy}dA}$ where $R = \lbrack 0,1\rbrack\ \times \ \lbrack 0,\text{ln}\ 5\rbrack.$

Applications of Double Integrals 二重积分的应用

Double integrals are very useful for finding the area of a region bounded by curves of functions. We describe this situation in more detail in the next section. However, if the region is a rectangular shape, we can find its area by integrating the constant function $f(x,y) = 1$ over the region $R.$

The area of the region $R$ is given by $A(R) = {\iint\limits_{R}{1dA}}.$

This definition makes sense because using $f(x,y) = 1$ and evaluating the integral make it a product of length and width. Let’s check this formula with an example and see how this works.

Finding Area Using a Double Integral 用二重积分求面积

Find the area of the region $R = \left\{ \left( {x,y} \right) \middle| 0 \leq x \leq 3,0 \leq y \leq 2 \right\}$ by using a double integral, that is, by integrating 1 over the region $R.$

Solution 解

The region is rectangular with length 3 and width 2, so we know that the area is 6. We get the same answer when we use a double integral:

$$A(R) = {\int\limits_{0}^{2}\ {\int\limits_{0}^{3}{1dx\ dy}}} = {\int\limits_{0}^{2}\left\lbrack \left. x \right|_{0}^{3} \right\rbrack}dy = {\int\limits_{0}^{2}{3dy = 3{\int\limits_{0}^{2}{dy = 3\left. y \right|_{0}^{2}}}}} = 3(2) = 6.$$

We have already seen how double integrals can be used to find the volume of a solid bounded above by a function $f(x,y)$ over a region $R$ provided $f(x,y) \geq 0$ for all $(x,y)$ in $R.$ Here is another example to illustrate this concept.

Volume of an Elliptic Paraboloid 椭球抛物面的体积

Find the volume $V$ of the solid $S$ that is bounded by the elliptic paraboloid $2x^{2} + y^{2} + z = 27,$ the planes $x = 3$ and $y = 3,$ and the three coordinate planes.

Solution 解

First notice the graph of the surface $z = 27 - 2x^{2} - y^{2}$ in Figure 5.9(a) and above the square region $R_{1} = \lbrack-3,3\rbrack\ \times \ \lbrack-3,3\rbrack.$ However, we need the volume of the solid bounded by the elliptic paraboloid $2x^{2} + y^{2} + z = 27,$ the planes $x = 3$ and $y = 3,$ and the three coordinate planes.

Now let’s look at the graph of the surface in Figure 5.9(b). We determine the volume *V* by evaluating the double integral over $R_{2}\text{:}$

$$\begin{array}{clccc} V & {= {\iint\limits_{R}{z\ dA =}}{\iint\limits_{R}{\left( {27 - 2x^{2} - y^{2}} \right)dA}}} & & & \\ & {= {\int\limits_{y = 0}^{y = 3}\ {\int\limits_{x = 0}^{x = 3}{\left( {27 - 2x^{2} - y^{2}} \right)dx\ dy}}}} & & & \text{Convert to iterated integral.} \\ & {= {\int\limits_{y = 0}^{y = 3}\left. \left\lbrack {27x - \frac{2}{3}x^{3} - y^{2}x} \right\rbrack \right|_{x = 0}^{x = 3}}dy} & & & {\text{Integrate with respect to}\ x.} \\ & {= {\int\limits_{y = 0}^{y = 3}{\left( {63 - 3y^{2}} \right)dy}} = 63y - \left. y^{3} \right|_{y = 0}^{y = 3} = 162.} & & & \end{array}$$

Find the volume of the solid bounded above by the graph of $f(x,y) = xy\ \text{sin}(x^{2}y)$ and below by the $xy$-plane on the rectangular region $R = \lbrack 0,1\rbrack\ \times \ \lbrack 0,\pi\rbrack.$

Recall that we defined the average value of a function of one variable on an interval $\lbrack a,b\rbrack$ as

$$f_{\text{ave}} = \frac{1}{b - a}{\int\limits_{a}^{b}{f(x)dx}}.$$

Similarly, we can define the average value of a function of two variables over a region *R*. The main difference is that we divide by an area instead of the width of an interval.

The average value of a function of two variables over a region $R$ is

$$f_{\text{ave}} = \frac{1}{\text{Area}\ R}{\iint\limits_{R}{f(x,y)dA.}}$$ (5.4)

In the next example we find the average value of a function over a rectangular region. This is a good example of obtaining useful information for an integration by making individual measurements over a grid, instead of trying to find an algebraic expression for a function.

Calculating Average Storm Rainfall 计算风暴平均降雨量

The weather map in Figure 5.10 shows an unusually moist storm system associated with the remnants of Hurricane Karl, which dumped 4–8 inches (100–200 mm) of rain in some parts of the Midwest on September 22–23, 2010. The area of rainfall measured 300 miles east to west and 250 miles north to south. Estimate the average rainfall over the entire area in those two days.

Solution 解

Place the origin at the southwest corner of the map so that all the values can be considered as being in the first quadrant and hence all are positive. Now divide the entire map into six rectangles $\left( {m = 3\ \text{and}\ n = 2} \right),$ as shown in Figure 5.11. Assume $f(x,y)$ denotes the storm rainfall in inches at a point approximately $x$ miles to the east of the origin and *y* miles to the north of the origin. Let $R$ represent the entire area of $250\ \times \ 300 = 75000$ square miles. Then the area of each subrectangle is

$$\text{Δ}A = \frac{1}{6}(75,000) = 12,500.$$

Assume $(x_{ij}^{*},y_{ij}^{*})$ are approximately the midpoints of each subrectangle $R_{ij}.$ Note the color-coded region at each of these points, and estimate the rainfall. The rainfall at each of these points can be estimated as:

At $\left( x*_{11},y*_{11} \right)$ the rainfall is 0.08.

At $\left( x*_{21},y*_{21} \right)$ the rainfall is 0.08.

At $\left( x*_{31},y*_{31} \right)$ the rainfall is 0.01.

At $\left( x*_{12},y*_{12} \right)$ the rainfall is 1.70.

At $\left( x*_{22},y*_{22} \right)$ the rainfall is 1.74.

At $\left( x*_{32},y*_{32} \right)$ the rainfall is 3.00.

According to our definition, the average storm rainfall in the entire area during those two days was

$$\begin{matrix} f_{\text{ave}} & {= \frac{1}{\text{Area}\ R}\iint\limits_{R}{f(x,y)dx\ dy} = \frac{1}{75000}\iint\limits_{R}{f(x,y)dx\ dy}} \\ & {\approx \frac{1}{75,000}\sum\limits_{i = 1}^{3}\sum\limits_{j = 1}^{2}{f\left( x_{ij}^{*},y_{ij}^{*} \right)\text{Δ}A}} \\ & {\approx \frac{1}{75,000}\left\lbrack {f\left( x_{11}^{*},y_{11}^{*} \right)\text{Δ}A + f\left( x_{21}^{*},y_{21}^{*} \right)\text{Δ}A} \right.} \\ & {\quad\left. {+ f\left( x_{31}^{*},y_{31}^{*} \right)\text{Δ}A + f\left( x_{12}^{*},y_{12}^{*} \right)\text{Δ}A + f\left( x_{22}^{*},y_{22}^{*} \right)\text{Δ}A + f\left( x_{32}^{*},y_{32}^{*} \right)\text{Δ}A} \right\rbrack} \\ & {\approx \frac{1}{75,000}\lbrack 0.08 + 0.08 + 0.01 + 1.70 + 1.74 + 3.00\rbrack\text{Δ}A} \\ & {\approx \frac{1}{75,000}\lbrack 0.08 + 0.08 + 0.01 + 1.70 + 1.74 + 3.00\rbrack 12500} \\ & {\approx \frac{5}{30}\lbrack 0.08 + 0.08 + 0.01 + 1.70 + 1.74 + 3.00\rbrack} \\ & {\approx 1.10.} \end{matrix}$$

During September 22–23, 2010 this area had an average storm rainfall of approximately 1.10 inches.

A contour map is shown for a function $f(x,y)$ on the rectangle $R = \lbrack-3,6\rbrack\ \times \ \lbrack-1,4\rbrack.$

1. Use the midpoint rule with $m = 3$ and $n = 2$ to estimate the value of ${\iint\limits_{R}{f(x,y)dA}}.$

2. Estimate the average value of the function $f(x,y).$

Section 5.1 Exercises 5.1 节习题

In the following exercises, use the midpoint rule with $m = 4$ and $n = 2$ to estimate the volume of the solid bounded by the surface $z = f(x,y),$ the vertical planes $x = 1,$ $x = 2,$ $y = 1,$ and $y = 2,$ and the horizontal plane $z = 0.$

1.

$f(x,y) = 4x + 2y + 8xy$

2\.

$f(x,y) = 16x^{2} + \frac{y}{2}$

In the following exercises, estimate $\iint_{R}f\left( {x,y} \right)dA$ by using a Riemann sum with $m = n = 2$ and the sample points to be the lower left corners of the subrectangles of the partition.

3.

$f(x,y) = \text{sin}\ x - \text{cos}\ y,$ $R = \lbrack 0,\pi\rbrack\ \times \ \lbrack 0,\pi\rbrack$

4\.

$f(x,y) = \text{cos}\ x + \text{cos}\ y,$ $R = \left\lbrack {0,\pi} \right\rbrack\ \times \ \left\lbrack {0,\frac{\pi}{2}} \right\rbrack$

5.

Use the midpoint rule with $m = n = 2$ to estimate ${\iint\limits_{R}{f(x,y)dA}},$ where the values of the function *f* on $R = \lbrack 8,10\rbrack\ \times \ \lbrack 9,11\rbrack$ are given in the following table.

| | | | | | |

|---------|---------|-----|-----|------|-----|

| | *y* | | | | |

| *x* | 9 | 9.5 | 10 | 10.5 | 11 |

| 8 | 9.8 | 5 | 6.7 | 5 | 5.6 |

| 8.5 | 9.4 | 4.5 | 8 | 5.4 | 3.4 |

| 9 | 8.7 | 4.6 | 6 | 5.5 | 3.4 |

| 9.5 | 6.7 | 6 | 4.5 | 5.4 | 6.7 |

| 10 | 6.8 | 6.4 | 5.5 | 5.7 | 6.8 |

6\.

The values of the function *f* on the rectangle $R = \lbrack 0,2\rbrack\ \times \ \lbrack 7,9\rbrack$ are given in the following table. Estimate the double integral $\iint\limits_{R}{f(x,y)dA}$ by using a Riemann sum with $m = n = 2.$ Select the sample points to be the upper right corners of the subsquares of *R*.

| | | | |

|-------------|-------------|-------------|-------------|

| | $y_{0} = 7$ | $y_{1} = 8$ | $y_{2} = 9$ |

| $x_{0} = 0$ | 10.22 | 10.21 | 9.85 |

| $x_{1} = 1$ | 6.73 | 9.75 | 9.63 |

| $x_{2} = 2$ | 5.62 | 7.83 | 8.21 |

7.

The depth of a children’s 4-ft by 4-ft swimming pool, measured at 1-ft intervals, is given in the following table.

1. Estimate the volume of water in the swimming pool by using a Riemann sum with $m = n = 2.$ Select the sample points using the midpoint rule on $R = \lbrack 0,4\rbrack\ \times \ \lbrack 0,4\rbrack.$

2. Approximate the average depth of the swimming pool.

| | | | | | |

|---------|---------|-----|-----|-----|-----|

| | *y* | | | | |

| *x* | 0 | 1 | 2 | 3 | 4 |

| 0 | 1 | 1.5 | 2 | 2.5 | 3 |

| 1 | 1 | 1.5 | 2 | 2.5 | 3 |

| 2 | 1 | 1.5 | 1.5 | 2.5 | 3 |

| 3 | 1 | 1 | 1.5 | 2 | 2.5 |

| 4 | 1 | 1 | 1 | 1.5 | 2 |

8\.

The depth of a 3-ft by 3-ft hole in the ground, measured at 1-ft intervals, is given in the following table.

1. Estimate the volume of the hole by using a Riemann sum with $m = n = 3$ and the sample points to be the upper left corners of the subsquares of *R*.

2. Approximate the average depth of the hole.

| | | | | |

|---------|---------|-----|-----|-----|

| | *y* | | | |

| *x* | 0 | 1 | 2 | 3 |

| 0 | 6 | 6.5 | 6.4 | 6 |

| 1 | 6.5 | 7 | 7.5 | 6.5 |

| 2 | 6.5 | 6.7 | 6.5 | 6 |

| 3 | 6 | 6.5 | 5 | 5.6 |

9.

The level curves $f(x,y) = k$ of the function *f* are given in the following graph, where *k* is a constant.

1. Apply the midpoint rule with $m = n = 2$ to estimate the double integral ${\iint\limits_{R}{f(x,y)dA}},$ where $R = \lbrack 0.2,1\rbrack\ \times \ \lbrack 0,0.8\rbrack.$

2. Estimate the average value of the function *f* on *R*.

10\.

The level curves $f(x,y) = k$ of the function *f* are given in the following graph, where *k* is a constant.

1. Apply the midpoint rule with $m = n = 2$ to estimate the double integral ${\iint\limits_{R}{f(x,y)dA}},$ where $R = \lbrack 0.1,0.5\rbrack\ \times \ \lbrack 0.1,0.5\rbrack.$

2. Estimate the average value of the function *f* on *R*.

11.

The solid lying under the surface $z = \sqrt{4 - y^{2}}$ and above the rectangular region $R = \lbrack 0,2\rbrack\ \times \ \left\lbrack {0,2} \right\rbrack$ is illustrated in the following graph. Evaluate the double integral ${\iint\limits_{R}{f(x,y)dA}},$ where $f(x,y) = \sqrt{4 - y^{2}},$ by finding the volume of the corresponding solid.

12\.

The solid lying under the plane $z = y + 4$ and above the rectangular region $R = \lbrack 0,2\rbrack\ \times \ \left\lbrack {0,4} \right\rbrack$ is illustrated in the following graph. Evaluate the double integral ${\iint\limits_{R}{f(x,y)dA}},$ where $f(x,y) = y + 4,$ by finding the volume of the corresponding solid.

In the following exercises, calculate the integrals by interchanging the order of integration.

13.

${\int\limits_{-1}^{1}\left( {\int\limits_{-2}^{2}{\left( {2x + 3y + 5} \right)dx}} \right)}dy$

14\.

${\int\limits_{0}^{2}\left( {\int\limits_{0}^{1}{\left( {x + 2e^{y} - 3} \right)dx}} \right)}dy$

15.

$\int\limits_{1}^{27}{\left( {{\int\limits_{1}^{2}\left( {\sqrt[3]{x} + \sqrt[3]{y}} \right)}dy} \right)dx}$

16\.

$\int\limits_{1}^{16}{\left( {{\int\limits_{1}^{8}\left( {\sqrt[4]{x} + 2\sqrt[3]{y}} \right)}dy} \right)dx}$

17.

${\int\limits_{\text{ln}\ 2}^{\text{ln}\ 3}\left( {\int\limits_{0}^{l}{e^{x + y}dy}} \right)}dx$

18\.

${\int\limits_{0}^{2}\left( {\int\limits_{0}^{1}{3^{x + y}dy}} \right)}dx$

19.

${\int\limits_{1}^{6}\left( {{\int\limits_{2}^{9}\frac{\sqrt{y}}{x^{2}}}dy} \right)}dx$

20\.

${\int\limits_{1}^{9}\left( {{\int\limits_{4}^{2}\frac{\sqrt{x}}{y^{2}}}dy} \right)}dx$

In the following exercises, evaluate the iterated integrals by choosing the order of integration.

21.

${\int\limits_{0}^{\pi}\ {\int\limits_{0}^{\pi\text{/}2}{\text{sin}(2x)\text{cos}(3y)}}}dx\ dy$

22\.

${\int\limits_{\pi\text{/}12}^{\pi\text{/}8}\ {\int\limits_{\pi\text{/}4}^{\pi\text{/}3}\left\lbrack {\text{cot}\ x + \text{tan}(2y)} \right\rbrack}}dx\ dy$

23.

$\int\limits_{1}^{e}\ {\int\limits_{1}^{e}{\left\lbrack {\frac{1}{x}\text{sin}(\text{ln}\ x) + \frac{1}{y}\text{cos}(\text{ln}\ y)} \right\rbrack dx\ dy}}$

24\.

$\int\limits_{1}^{e}\ {\int\limits_{1}^{e}{\frac{\text{sin}(\text{ln}\ x)\text{cos}(\text{ln}\ y)}{xy}dx\ dy}}$

25.

${\int\limits_{1}^{2}\ {\int\limits_{1}^{2}\left( {\frac{\text{ln}\ y}{x} + \frac{x}{2y + 1}} \right)}}dy\ dx$

26\.

${\int\limits_{1}^{e}\ {\int\limits_{1}^{2}{x^{2}\text{ln}(x)}}}dy\ dx$

27.

${\int\limits_{1}^{\sqrt{3}}\ {\int\limits_{1}^{2}{y\ \text{arctan}\left( \frac{1}{x} \right)}}}dy\ dx$

28\.

${\int\limits_{0}^{1}\ {\int\limits_{0}^{1\text{/}2}\left( {\text{arcsin}\ x + \text{arcsin}\ y} \right)}}dy\ dx$

29.

${\int\limits_{0}^{1}\ {\int\limits_{1}^{2}{xe^{x + 4y}}}}dy\ dx$

30\.

${\int\limits_{1}^{2}\ {\int\limits_{0}^{1}{xe^{x - y}}}}dy\ dx$

31.

${\int\limits_{1}^{e}\ {\int\limits_{1}^{e}\left( {\frac{\text{ln}\ y}{\sqrt{y}} + \frac{\text{ln}\ x}{\sqrt{x}}} \right)}}dy\ dx$

32\.

${\int\limits_{1}^{e}\ {\int\limits_{1}^{e}\left( {\frac{x\ \text{ln}\ y}{\sqrt{y}} + \frac{y\ \text{ln}\ x}{\sqrt{x}}} \right)}}dy\ dx$

33.

${\int\limits_{0}^{1}\ {\int\limits_{1}^{2}\left( \frac{x}{x^{2} + y^{2}} \right)}}dy\ dx$

34\.

${\int\limits_{0}^{1}\ {\int\limits_{1}^{2}\frac{y}{x + y^{2}}}}dy\ dx$

In the following exercises, find the average value of the function over the given rectangles.

35.

$f(x,y) = \text{−}x + 2y,$ $R = \lbrack 0,1\rbrack\ \times \ \left\lbrack {0,1} \right\rbrack$

36\.

$f(x,y) = x^{4} + 2y^{3},$ $R = \lbrack 1,2\rbrack\ \times \ \left\lbrack {2,3} \right\rbrack$

37.

$f(x,y) = \text{sinh}\ x + \text{sinh}\ y,$ $R = \lbrack 0,1\rbrack\ \times \ \left\lbrack {0,2} \right\rbrack$

38\.

$f(x,y) = \text{arctan}(xy),$ $R = \left\lbrack {0,1} \right\rbrack\ \times \ \left\lbrack {0,1} \right\rbrack$

39.

Let *f* and *g* be two continuous functions such that $0 \leq m_{1} \leq f(x) \leq M_{1}$ for any $x \in \lbrack a,b\rbrack$ and $0 \leq m_{2} \leq g(y) \leq M_{2}$ for any $y \in \lbrack c,d\rbrack.$ Show that the following inequality is true:

$m_{1}m_{2}(b - a)(d - c) \leq \int\limits_{a}^{b}\ \int\limits_{c}^{d}{f(x)g(y)dy\ dx} \leq M_{1}M_{2}(b - a)(d - c).$

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

40\.

$\frac{1}{e^{2}} \leq {\iint\limits_{R}{e^{\text{−}x^{2} - y^{2}}dA}} \leq 1,$ where $R = \left\lbrack {0,1} \right\rbrack\ \times \ \left\lbrack {0,1} \right\rbrack$

41.

$\frac{\pi^{2}}{144} \leq {\iint\limits_{R}{\text{sin}\ x\ \text{cos}\ y\ dA}} \leq \frac{\pi^{2}}{48},$ where $R = \left\lbrack {\frac{\pi}{6},\frac{\pi}{3}} \right\rbrack\ \times \ \left\lbrack {\frac{\pi}{6},\frac{\pi}{3}} \right\rbrack$

42.

$0 \leq \iint\limits_{R}e^{\text{−}y}\text{cos}\ x\ dA \leq \left( \frac{\pi}{2} \right)^{2},$ where $R = \left\lbrack {0,\frac{\pi}{2}} \right\rbrack\ \times \ \left\lbrack {0,\frac{\pi}{2}} \right\rbrack$

43.

$0 \leq {\iint\limits_{R}{\left( {\text{ln}\ x} \right)\left( {\text{ln}\ y} \right)dA}} \leq \left( {e - 1} \right)^{2},$ where $R = \left\lbrack {1,e} \right\rbrack\ \times \ \left\lbrack {1,e} \right\rbrack$

44.

Let *f* and *g* be two continuous functions such that $0 \leq m_{1} \leq f(x) \leq M_{1}$ for any $x \in \lbrack a,b\rbrack$ and $0 \leq m_{2} \leq g(y) \leq M_{2}$ for any $y \in \lbrack c,d\rbrack.$ Show that the following inequality is true:

$\left( {m_{1} + m_{2}} \right)(b - a)(d - c) \leq {\int\limits_{a}^{b}\ {\int\limits_{c}^{d}{\left\lbrack {f(x) + g(y)} \right\rbrack dy\ dx}}} \leq \left( {M_{1} + M_{2}} \right)(b - a)(d - c).$

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

45.

$\frac{2}{e} \leq {\iint\limits_{R}{\left( {e^{\text{−}x^{2}} + e^{\text{−}y^{2}}} \right)dA}} \leq 2,$ where $R = \left\lbrack {0,1} \right\rbrack\ \times \ \left\lbrack {0,1} \right\rbrack$

46.

$\frac{\pi^{2}}{36} \leq {\iint\limits_{R}\left( {\text{sin}\ x + \text{cos}\ y} \right)}dA \leq \frac{\pi^{2}\sqrt{3}}{36},$ where $R = \left\lbrack {\frac{\pi}{6},\frac{\pi}{3}} \right\rbrack\ \times \ \left\lbrack {\frac{\pi}{6},\frac{\pi}{3}} \right\rbrack$

47.

$\frac{\pi^{2}}{4}e^{\frac{- \pi}{2}} \leq \iint\limits_{R}\left( \text{cos}\ x + e^{- y} \right)dA \leq \frac{\pi^{2}}{2},$ where $R = \left\lbrack {0,\frac{\pi}{2}} \right\rbrack\ \times \ \left\lbrack {0,\frac{\pi}{2}} \right\rbrack$

48.

$0 \leq \iint\limits_{R}\left( \text{ln}~x + \text{ln}\ y \right)dA \leq 2\left( {e - 1} \right)^{2},$ where $R = \left\lbrack {1,e} \right\rbrack\ \times \ \left\lbrack {1,e} \right\rbrack$

In the following exercises, the function *f* is given in terms of double integrals.

1. Determine the explicit form of the function *f*.

2. Find the volume of the solid under the surface $z = f(x,y)$ and above the region *R*.

3. Find the average value of the function *f* on *R*.

4. Use a computer algebra system (CAS) to plot $z = f(x,y)$ and $z = f_{\text{ave}}$ in the same system of coordinates.

49.

\[T\] $f(x,y) = {\int\limits_{0}^{y}\ {\int\limits_{0}^{x}{(xs + yt)ds\ dt}}},$ where $(x,y) \in R = \lbrack 0,1\rbrack\ \times \ \lbrack 0,1\rbrack$

50\.

\[T\] $f(x,y) = {\int\limits_{0}^{x}\ {\int\limits_{0}^{y}{\left\lbrack {\text{cos}(s) + \text{cos}(t)} \right\rbrack dt\ ds}}},$ where $(x,y) \in R = \lbrack 0,3\rbrack\ \times \ \lbrack 0,3\rbrack$

51.

Show that if *f* and *g* are continuous on $\lbrack a,b\rbrack$ and $\lbrack c,d\rbrack,$ respectively, then

${\int\limits_{a}^{b}\ {\int\limits_{c}^{d}{\left\lbrack {f(x) + g(y)} \right\rbrack dy\ dx}}} = (d - c){\int\limits_{a}^{b}{f(x)dx}}$

$\mspace{90mu} + {\int\limits_{a}^{b}\ {\int\limits_{c}^{d}{g(y)dy}}}\ dx = (b - a){\int\limits_{c}^{d}{g(y)dy +}}{\int\limits_{c}^{d}\ {\int\limits_{a}^{b}{f(x)dx}}}\ dy.$

52.

Show that ${\int\limits_{a}^{b}\ {\int\limits_{c}^{d}{yf(x) + xg(y)dy\ dx}}} = \frac{1}{2}\left( {d^{2} - c^{2}} \right)\left( {\int\limits_{a}^{b}{f(x)dx}} \right) + \frac{1}{2}\left( {b^{2} - a^{2}} \right)\left( {\int\limits_{c}^{d}{g(y)dy}} \right).$

53.

\[T\] Consider the function $f(x,y) = e^{\text{−}x^{2} - y^{2}},$ where $(x,y) \in R = \lbrack-1,1\rbrack\ \times \ \lbrack-1,1\rbrack.$

1. Use the midpoint rule with $m = n = 2,4\text{,…,}\ 10$ to estimate the double integral $I = {\iint\limits_{R}{e^{\text{−}x^{2} - y^{2}}dA}}.$ Round your answers to the nearest hundredths.

2. For $m = n = 2,$ find the average value of *f* over the region *R*. Round your answer to the nearest hundredths.

3. Use a CAS to graph in the same coordinate system the solid whose volume is given by $\iint\limits_{R}{e^{\text{−}x^{2} - y^{2}}dA}$ and the plane $z = f_{\text{ave}}.$

54\.

\[T\] Consider the function $f(x,y) = \text{sin}\left( x^{2} \right)\text{cos}\left( y^{2} \right),$ where $(x,y) \in R = \lbrack-1,1\rbrack\ \times \ \lbrack-1,1\rbrack.$

1. Use the midpoint rule with $m = n = 2,4\text{,…,}\ 10$ to estimate the double integral $I = {\iint\limits_{R}{\text{sin}\left( x^{2} \right)\text{cos}\left( y^{2} \right)dA}}.$ Round your answers to the nearest hundredths.

2. For $m = n = 2,$ find the average value of *f* over the region *R.* Round your answer to the nearest hundredths.

3. Use a CAS to graph in the same coordinate system the solid whose volume is given by $\iint\limits_{R}{\text{sin}\left( x^{2} \right)\text{cos}\left( y^{2} \right)dA}$ and the plane $z = f_{\text{ave}}.$

In the following exercises, the functions $f_{n}$ are given, where $n \geq 1$ is a natural number.

1. Find the volume of the solids $S_{n}$ under the surfaces $z = f_{n}(x,y)$ and above the region *R*.

2. Determine the limit of the volumes of the solids $S_{n}$ as *n* increases without bound.

55.

$f_{n}(x,y) = x^{n} + y^{n} + xy,(x,y) \in R = \lbrack 0,1\rbrack\ \times \lbrack 0,1\rbrack$

56\.

$f_{n}(x,y) = \frac{1}{x^{n}} + \frac{1}{y^{n}},(x,y) \in R = \lbrack 1,2\rbrack\ \times \lbrack 1,2\rbrack$

57.

Show that the average value of a function *f* on a rectangular region $R = \lbrack a,b\rbrack\ \times \ \lbrack c,d\rbrack$ is $f_{\text{ave}} \approx \frac{1}{mn}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f\left( {x_{ij}^{*},y_{ij}^{*}} \right)}}},$ where $\left( {x_{ij}^{*},y_{ij}^{*}} \right)$ are the sample points of the partition of *R*, where $1 \leq i \leq m$ and $1 \leq j \leq n.$

58.

Use the midpoint rule with $m = n$ to show that the average value of a function *f* on a rectangular region $R = \lbrack a,b\rbrack\ \times \ \lbrack c,d\rbrack$ is approximated by

$$f_{\text{ave}} \approx \frac{1}{n^{2}}{\sum\limits_{i,j = 1}^{n}{f\left( {\frac{1}{2}\left( {x_{i - 1} + x_{i}} \right),\frac{1}{2}\left( {y_{j - 1} + y_{j}} \right)} \right)}}.$$ 59.

An isotherm map is a chart connecting points having the same temperature at a given time for a given period of time. Use the preceding exercise and apply the midpoint rule with $m = n = 2$ to find the average temperature over the region given in the following figure.

---

5.2 Double Integrals over General Regions 5.2 一般区域上的二重积分

• 5.2.1 Recognize when a function of two variables is integrable over a general region.
• 5.2.2 Evaluate a double integral by computing an iterated integral over a region bounded by two vertical lines and two functions of $x,$ or two horizontal lines and two functions of $y.$
• 5.2.3 Simplify the calculation of an iterated integral by changing the order of integration.
• 5.2.4 Use double integrals to calculate the volume of a region between two surfaces or the area of a plane region.
• 5.2.5 Solve problems involving double improper integrals.

• 5.2.1 识别二元函数在一般区域上何时可积。
• 5.2.2 通过把二重积分化为累次积分来计算，其中积分区域由两条竖直线与两条关于 $x$ 的函数、或两条水平线与两条关于 $y$ 的函数所围成。
• 5.2.3 通过改变积分次序简化累次积分的计算。
• 5.2.4 利用二重积分计算两曲面之间区域的体积或平面区域的面积。
• 5.2.5 求解涉及二重反常积分的问题。

In Double Integrals over Rectangular Regions, we studied the concept of double integrals and examined the tools needed to compute them. We learned techniques and properties to integrate functions of two variables over rectangular regions. We also discussed several applications, such as finding the volume bounded above by a function over a rectangular region, finding area by integration, and calculating the average value of a function of two variables.

In this section we consider double integrals of functions defined over a general bounded region $D$ on the plane. Most of the previous results hold in this situation as well, but some techniques need to be extended to cover this more general case.

General Regions of Integration 一般区域上的积分

An example of a general bounded region $D$ on a plane is shown in Figure 5.12. Since $D$ is bounded on the plane, there must exist a rectangular region $R$ on the same plane that encloses the region $D,$ that is, a rectangular region $R$ exists such that $D$ is a subset of $R\left( {D \subseteq R} \right).$

Suppose $z = f\left( {x,y} \right)$ is defined on a general planar bounded region $D$ as in Figure 5.12. In order to develop double integrals of $f$ over $D,$ we extend the definition of the function to include all points on the rectangular region $R$ and then use the concepts and tools from the preceding section. But how do we extend the definition of $f$ to include all the points on $R?$ We do this by defining a new function $g\left( {x,y} \right)$ on $R$ as follows:

$$g\left( {x,y} \right) = \begin{cases} {f\left( {x,y} \right)} & {\text{if}\ \left( {x,y} \right)\ \text{is in}\ D} \\ 0 & {\text{if}\ \left( {x,y} \right)\ \text{is in}\ R\ \text{but not in}\ D} \end{cases}$$

Note that we might have some technical difficulties if the boundary of $D$ is complicated. So we assume the boundary to be a piecewise smooth and continuous simple closed curve. Also, since all the results developed in Double Integrals over Rectangular Regions used an integrable function $f\left( {x,y} \right),$ we must be careful about $g\left( {x,y} \right)$ and verify that $g\left( {x,y} \right)$ is an integrable function over the rectangular region $R.$ This happens as long as the region $D$ is bounded by simple closed curves. For now we will concentrate on the descriptions of the regions rather than the function and extend our theory appropriately for integration.

We consider two types of planar bounded regions.

A region $D$ in the $\left( {x,y} \right)$-plane is of Type I if it lies between two vertical lines and the graphs of two continuous functions $g_{1}(x)$ and $g_{2}(x).$ That is (Figure 5.13),

$$D = \left\{ {\left. \left( {x,y} \right) \right|a \leq x \leq b,g_{1}(x) \leq y \leq g_{2}(x)} \right\}.$$

A region $D$ in the $xy$ plane is of Type II if it lies between two horizontal lines and the graphs of two continuous functions $h_{1}(y)\ \text{and}\ h_{2}(y).$ That is (Figure 5.14),

$$D = \left\{ {\left. \left( {x,y} \right) \right|c \leq y \leq d,h_{1}(y) \leq x \leq h_{2}(y)} \right\}.$$

Describing a Region as Type I and Also as Type II 将区域同时描述为Ⅰ型和Ⅱ型

Consider the region in the first quadrant between the functions $y = \sqrt{x}$ and $y = x^{3}$ (Figure 5.15). Describe the region first as Type I and then as Type II.

Solution 解

When describing a region as Type I, we need to identify the function that lies above the region and the function that lies below the region. Here, region $D$ is bounded above by $y = \sqrt{x}$ and below by $y = x^{3}$ in the interval for $x\ \text{in}\ \left\lbrack {0,1} \right\rbrack.$ Hence, as Type I, $D$ is described as the set $\left\{ {\left. \left( {x,y} \right) \right|0 \leq x \leq 1,x^{3} \leq y \leq \sqrt{x}} \right\}.$

However, when describing a region as Type II, we need to identify the function that lies on the left of the region and the function that lies on the right of the region. Here, the region $D$ is bounded on the left by $x = y^{2}$ and on the right by $x = \sqrt[3]{y}$ in the interval for *y* in $\lbrack 0,1\rbrack.$ Hence, as Type II, $D$ is described as the set $\left\{ {\left. \left( {x,y} \right) \right|0 \leq y \leq 1,y^{2} \leq x \leq \sqrt[3]{y}} \right\}.$

Consider the region in the first quadrant between the functions $y = 2x$ and $y = x^{2}.$ Describe the region first as Type I and then as Type II.

Double Integrals over Nonrectangular Regions 一般区域上的二重积分

To develop the concept and tools for evaluation of a double integral over a general, nonrectangular region, we need to first understand the region and be able to express it as Type I or Type II or a combination of both. Without understanding the regions, we will not be able to decide the limits of integrations in double integrals. As a first step, let us look at the following theorem.

Double Integrals over Nonrectangular Regions 一般区域上的二重积分

Suppose $g\left( {x,y} \right)$ is the extension to the rectangle $R$ of the integrable function $f\left( {x,y} \right)$ defined on the region $D$, where $D$ is inside $R$. Sample regions are as shown in Figure 5.12. Then $g\left( {x,y} \right)$ is integrable and we define the double integral of $f\left( {x,y} \right)$ over $D$ by

$${\iint\limits_{D}{f\left( {x,y} \right)}}dA = {\iint\limits_{R}{g\left( {x,y} \right)}}dA.$$

The right-hand side of this equation is what we have seen before, so this theorem is reasonable because $R$ is a rectangle and ${\iint\limits_{R}{g\left( {x,y} \right)}}dA$ has been discussed in the preceding section. Also, the equality works because the values of $g\left( {x,y} \right)$ are $0$ for any point $\left( {x,y} \right)$ that lies outside $D,$ and hence these points do not add anything to the integral. However, it is important that the rectangle $R$ contains the region $D.$

As a matter of fact, if the region $D$ is bounded by smooth curves on a plane and we are able to describe it as Type I or Type II or a mix of both, then we can use the following theorem and not have to find a rectangle $R$ containing the region.

Fubini’s Theorem (Strong Form) 富比尼定理（强形式）

For a function $f\left( {x,y} \right)$ that is continuous on a region $D$ of Type I, we have

$${\iint\limits_{D}{f(x,y)dA}} = {\iint\limits_{D}{f(x,y)dy\ dx}} = {\int\limits_{a}^{b}{\left\lbrack {\int\limits_{g_{1}(x)}^{g_{2}(x)}{f(x,y)dy}} \right\rbrack dx}}.$$ (5.5)

Similarly, for a function $f\left( {x,y} \right)$ that is continuous on a region $D$ of Type II, we have

$${\iint\limits_{D}{f(x,y)dA}} = {\iint\limits_{D}{f(x,y)dx\ dy}} = {\int\limits_{c}^{d}{\left\lbrack {\int\limits_{h_{1}(y)}^{h_{2}(y)}{f(x,y)dx}} \right\rbrack dy}}.$$ (5.6)

The integral in each of these expressions is an iterated integral, similar to those we have seen before. Notice that, in the inner integral in the first expression, we integrate $f\left( {x,y} \right)$ with $x$ being held constant and the limits of integration being $g_{1}(x)\ \text{and}\ g_{2}(x).$ In the inner integral in the second expression, we integrate $f\left( {x,y} \right)$ with $y$ being held constant and the limits of integration are $h_{1}(y)\ \text{and}\ h_{2}(y).$

Evaluating an Iterated Integral over a Type I Region 在Ⅰ型区域上计算累次积分

Evaluate the integral $\iint\limits_{D}{x^{2}e^{xy}dA}$ where $D$ is shown in Figure 5.16.

Solution 解

First construct the region $D$ as a Type I region (Figure 5.16). Here $D = \left\{ {\left. \left( {x,y} \right) \right|0 \leq x \leq 2,\frac{1}{2}x \leq y \leq 1} \right\}.$ Then we have

$${\iint\limits_{D}{x^{2}e^{xy}dA}} = {\int\limits_{x = 0}^{x = 2}\ {\int\limits_{y = {1\text{/}2}x}^{y = 1}{x^{2}e^{xy}dy\ dx}}}.$$

Therefore, we have

$$\begin{array}{clccc} {\int\limits_{x = 0}^{x = 2}\ {\int\limits_{y = \frac{1}{2}x}^{y = 1}{x^{2}e^{xy}dy\ dx}}} & {= {\int\limits_{x = 0}^{x = 2}\left\lbrack {\int\limits_{y = {1\text{/}2}x}^{y = 1}{x^{2}e^{xy}dy}} \right\rbrack}dx} & & & \text{Iterated integral for a Type I region.} \\ & {= {\int\limits_{x = 0}^{x = 2}\left. \left\lbrack {x^{2}\frac{e^{xy}}{x}} \right\rbrack \right|_{y = {1\text{/}2}x}^{y = 1}}dx} & & & \begin{array}{l} {\text{Integrate with respect to}\ y\ \text{using}} \\ {u\text{-substitution with}\ u = xy\ \text{where}\ x\ \text{is held}} \\ \text{constant.} \end{array} \\ & {= {\int\limits_{x = 0}^{x = 2}\left\lbrack {xe^{x} - xe^{x^{2}\text{/}2}} \right\rbrack}dx} & & & \begin{array}{l} {\text{Integrate with respect to}\ x\ \text{using}} \\ {u\text{-substitution with}\ u = \frac{1}{2}x^{2}.} \end{array} \\ & {= \left. \left\lbrack {xe^{x} - e^{x} - e^{\frac{1}{2}x^{2}}} \right\rbrack \right|_{x = 0}^{x = 2} = 2} & & & \end{array}$$

In Example 5.12, we could have looked at the region in another way, such as $D = \left\{ {\left. \left( {x,y} \right) \right|0 \leq y \leq 1,0 \leq x \leq 2y} \right\}$ (Figure 5.17).

This is a Type II region and the integral would then look like

$${\iint\limits_{D}{x^{2}e^{xy}dA}} = {\int\limits_{y = 0}^{y = 1}\ {\int\limits_{x = 0}^{x = 2y}{x^{2}e^{xy}dx\ dy.}}}$$

However, if we integrate first with respect to $x,$ this integral is lengthy to compute because we have to use integration by parts twice.

Evaluating an Iterated Integral over a Type II Region 在Ⅱ型区域上计算累次积分

Evaluate the integral $\iint\limits_{D}{\left( {3x^{2} + y^{2}} \right)dA}$ where $= \left\{ {\left. \left( {x,y} \right) \right| - 2 \leq y \leq 3,y^{2} - 3 \leq x \leq y + 3} \right\}.$

Solution 解

Notice that $D$ can be seen as either a Type I or a Type II region, as shown in Figure 5.18. However, in this case describing $D$ as Type $\text{I}$ is more complicated than describing it as Type II. Therefore, we use $D$ as a Type II region for the integration.

Choosing this order of integration, we have

$$\begin{array}{clccc} {\iint\limits_{D}{\left( {3x^{2} + y^{2}} \right)dA}} & {= {\int\limits_{y = -2}^{y = 3}\ {\int\limits_{x = y^{2} - 3}^{x = y + 3}{\left( {3x^{2} + y^{2}} \right)dx\ dy}}}} & & & \text{Iterated integral, Type II region.} \\ & {= \left. {\int\limits_{y = -2}^{y = 3}\left( {x^{3} + xy^{2}} \right)} \right|_{y^{2} - 3}^{y + 3}dy} & & & {\text{Integrate with respect to}\ x.} \\ & {= {\int\limits_{y = -2}^{y = 3}\left( {\left( {y + 3} \right)^{3} + \left( {y + 3} \right)y^{2} - \left( {y^{2} - 3} \right)^{3} - \left( {y^{2} - 3} \right)y^{2}} \right)}dy} & & & \\ & {= {\int\limits_{-2}^{3}\left( {54 + 27y - 12y^{2} + 2y^{3} + 8y^{4} - y^{6}} \right)}dy} & & & {\text{Integrate with respect to}\ y.} \\ & {= \left. \left\lbrack {54y + \frac{27y^{2}}{2} - 4y^{3} + \frac{y^{4}}{2} + \frac{8y^{5}}{5} - \frac{y^{7}}{7}} \right\rbrack \right|_{-2}^{3}} & & & \\ & {= \frac{2375}{7}.} & & & \end{array}$$

Section 5.2 Exercises 5.2 节习题

In the following exercises, specify whether the region is of Type I or Type II.

60\.

The region $D$ bounded by $y = x^{3},$ $y = x^{3} + 1,$ $x = 0,$ and $x = 1$ as given in the following figure.

61.

Find the average value of the function $f\left( {x,y} \right) = 3xy$ on the region graphed in the previous exercise.

62\.

Find the area of the region $D$ given in the previous exercise.

63.

The region $D$ bounded by $y = \text{sin}\ x,y = 1 + \text{sin}\ x,x = 0,\ \text{and}\ x = \frac{\pi}{2}$ as given in the following figure.

64\.

Find the average value of the function $f\left( {x,y} \right) = \text{cos}\ x$ on the region graphed in the previous exercise.

65.

Find the area of the region $D$ given in the previous exercise.

66\.

The region $D$ bounded by $x = y^{2} - 1$ and $x = \sqrt{1 - y^{2}}$ as given in the following figure.

67.

Find the volume of the solid under the graph of the function $f\left( {x,y} \right) = xy + 1$ and above the region in the figure in the previous exercise.

68\.

The region $D$ bounded by $y = 0,x = -10 + y,\ \text{and}\ x = 10 - y$ as given in the following figure.

69.

Find the signed volume of the solid under the graph of the function $f\left( {x,y} \right) = x + y$ and above the region in the figure from the previous exercise.

70\.

The region $D$ bounded by $y = 0,x = y - 1,$ $x = \frac{\pi}{2}$ as given in the following figure.

71.

The region $D$ bounded by $y = 0$ and $y = x^{2} - 1$ as given in the following figure.

72\.

Let $D$ be the region bounded by the curve $y = 2 - x^{2}$ and below the equations $y = x,y = \text{−}x,$ Explain why $D$ is neither of Type I nor II.

73.

Let $D$ be the region bounded above by the curve of the equation $y = 4 - x^{2}$ and below by $y = \text{cos}x$ and the $x$-axis. Explain why $D$ is neither of Type I nor II.

In the following exercises, evaluate the double integral ${\iint\limits_{D}{f\left( {x,y} \right)}}dA$ over the region $D.$

74\.

$f\left( {x,y} \right) = 2x + 5y$ and $D = \left\{ {\left. \left( {x,y} \right) \right|0 \leq x \leq 1,x^{3} \leq y \leq x^{3} + 1} \right\}$

75.

$f\left( {x,y} \right) = 1$ and $D = \left\{ {\left. \left( {x,y} \right) \right|0 \leq x \leq \frac{\pi}{2},\text{sin}\ x \leq y \leq 1 + \text{sin}\ x} \right\}$

76\.

$f\left( {x,y} \right) = 2$ and $D = \left\{ {\left. \left( {x,y} \right) \right|0 \leq y \leq 1,y - 1 \leq x \leq \text{arccos}\ y} \right\}$

77.

$f\left( {x,y} \right) = xy$ and $D = \left\{ {\left. \left( {x,y} \right) \right| - 1 \leq y \leq 1,y^{2} - 1 \leq x \leq \sqrt{1 - y^{2}}} \right\}$

78\.

$f\left( {x,y} \right) = \text{sin}\ y$ and $D$ is the triangular region with vertices $\left( {0,0} \right),\left( {0,3} \right),\ \text{and}\ \left( {3,0} \right)$

79.

$f\left( {x,y} \right) = \text{−}x + 1$ and $D$ is the triangular region with vertices $\left( {0,0} \right),\left( {0,2} \right),\ \text{and}\ \left( {2,2} \right)$

Evaluate the iterated integrals.

80\.

${\int\limits_{0}^{1}\ {\int\limits_{2x}^{3x}\left( {x + y^{2}} \right)}}dy\ dx$

81.

${\int\limits_{0}^{1}\ {\int\limits_{2\sqrt{x}}^{2\sqrt{x} + 1}\left( {xy + 1} \right)}}dy\ dx$

82\.

${\int\limits_{e}^{e^{2}}\ {\int\limits_{\text{ln}\ u}^{2}\left( {v + \text{ln}\ u} \right)}}dv\ du$

83.

${\int\limits_{1}^{2}\ {\int\limits_{\text{−}u^{2} - 1}^{\text{−}u}\left( {8uv} \right)}}dv\ du$

84\.

${\int\limits_{0}^{1}\ {\int\limits_{\text{−}\sqrt{1 - y^{2}}}^{\sqrt{1 - y^{2}}}\left( {2x + 4x^{3}} \right)}}dx\ dy$

85.

${\int\limits_{0}^{1\text{/}2}\ {\int\limits_{\text{−}\sqrt{1 - 4y^{2}}}^{\sqrt{1 - 4y^{2}}}4}}dx\ dy$

86\.

Let $D$ be the region in the first quadrant bounded by $y = 1 - x^{2},y = 4 - x^{2},$ and the $x$- and $y$-axes.

1. Show that ${\iint\limits_{D}{x\ dA = {\int\limits_{0}^{1}\ {\int\limits_{1 - x^{2}}^{4 - x^{2}}{x\ dy\ dx}}}}} + {\int\limits_{1}^{2}\ {\int\limits_{0}^{4 - x^{2}}{x\ dy\ dx}}}$ by dividing the region $D$ into two regions of Type I.

2. Evaluate the integral ${\iint\limits_{D}{x\ dA}}.$

87.

Let $D$ be the region bounded by $y = 1,$ $y = x,$ $y = \text{ln}\ x,$ and the $x$-axis.

1. Show that $\iint\limits_{D}y\ dA = {\int\limits_{0}^{1}\ {\int\limits_{0}^{x}{y\ dy\ dx}}} + {\int\limits_{1}^{e}\ {\int\limits_{\text{ln}\ x}^{1}{y\ dy\ dx}}}$ by dividing $D$ into two regions of Type I.

2. Evaluate the integral ${\iint\limits_{D}{y\ dA}}.$

88\.

1. Show that $\iint\limits_{D}y^{2}dA = {\int\limits_{-1}^{0}\ {\int\limits_{\text{−}x}^{2 - x^{2}}{y^{2}dy\ dx}}} + {\int\limits_{0}^{1}\ {\int\limits_{x}^{2 - x^{2}}{y^{2}dy\ dx}}}$ by dividing the region $D$ into two regions of Type I, where $D = \left\{ (x,y) \middle| y \geq x,y \geq - x,y \leq 2 - x^{2} \right\}.$

2. Evaluate the integral ${\iint\limits_{D}y^{2}}dA.$

89.

Let $D$ be the region bounded by $y = x^{2},y = x + 2,$ and $y = \text{−}x.$

1. Show that $\iint\limits_{D}x\ dA = {\int\limits_{0}^{1}\ {\int\limits_{\text{−}y}^{\sqrt{y}}{x\ dx\ dy}}} + {\int\limits_{1}^{4}\ {\int\limits_{y - 2}^{\sqrt{y}}{x\ dx\ dy}}}$ by dividing the region $D$ into two regions of Type II, where $D = \left\{ (x,y) \middle| y \geq x^{2},y \geq - x,y \leq x + 2 \right\}.$

2. Evaluate the integral ${\iint\limits_{D}x}\ dA.$

90\.

The region $D$ bounded by $x = 0,y = x^{5} + 1,$ and $y = 3 - x^{2}$ is shown in the following figure. Find the area $A(D)$ of the region $D.$

91.

The region $D$ bounded by $y = \text{cos}\ x,y = 4 + \text{cos}\ x,$ and $x = \pm \frac{\pi}{3}$ is shown in the following figure. Find the area $A(D)$ of the region $D.$

92\.

Find the area $A(D)$ of the region $D = \left\{ (x,y) \middle| y \geq 1 - x^{2},y \leq 4 - x^{2},y \geq 0,x \geq 0 \right\}.$

93.

Let $D$ be the region bounded by $y = 1,y = x,y = \text{ln}\ x,$ and the $x$-axis. Find the area $A(D)$ of the region $D.$

94\.

Find the average value of the function $f\left( {x,y} \right) = \text{sin}\ y$ on the triangular region with vertices $\left( {0,0} \right),\left( {0,3} \right),$ and $\left( {3,0} \right).$

95.

Find the average value of the function $f\left( {x,y} \right) = \text{−}x + 1$ on the triangular region with vertices $\left( {0,0} \right),\left( {0,2} \right),$ and $\left( {2,2} \right).$

In the following exercises, change the order of integration and evaluate the integral.

96\.

$\int\limits_{-1}^{\pi\text{/}2}\ {\int\limits_{0}^{x + 1}{\text{sin}\ x\ dy\ dx}}$

97.

$\int\limits_{0}^{1}\ {\int\limits_{x - 1}^{1 - x}{x\ dy\ dx}}$

98\.

$\int\limits_{-1}^{0}\ {\int\limits_{\text{−}\sqrt{y + 1}}^{\sqrt{y + 1}}{y^{2}dx\ dy}}$

99.

$\int\limits_{\text{−1}}^{1}\ \int\limits_{\text{−}\sqrt{1–y^{2}}}^{\sqrt{1–y^{2}}}y\ dx\ dy$

100\.

The region $D$ is shown in the following figure. Evaluate the double integral $\iint\limits_{D}{\left( {x^{2} + y} \right)dA}$ by using the easier order of integration.

101.

The region $D$ is given in the following figure. Evaluate the double integral $\iint\limits_{D}{\left( {x^{2} - y^{2}} \right)dA}$ by using the easier order of integration.

102\.

Find the volume of the solid under the surface $z = 2x + y^{2}$ and above the region bounded by $y = x^{5}$ and $y = x.$

103.

Find the volume in the first octant of the solid under the plane $z = 3x + y$ and above the region determined by $y = x^{7}$ and $y = x.$

104\.

Find the volume of the solid under the plane $z = x - y$ and above the region bounded by $x = \text{tan}\ y,x = \text{−}\text{tan}\ y,$ and $x = 1.$

105.

Find the volume of the solid under the surface $z = x^{3}$ and above the plane region bounded by $x = \text{sin}\ y,x = \text{−}\text{sin}\ y,$ and $x = 1$ for values of $y$ between $y = \frac{\operatorname{–\pi}}{2}\ \text{and}\ y = \frac{\pi}{2}$

106\.

Let $g$ be a positive, increasing, and differentiable function on the interval $\left\lbrack {a,b} \right\rbrack.$ Show that the volume of the solid under the surface $z = g\prime(x)$ and above the region bounded by $y = 0,$ $y = g(x),$ $x = a,$ and $x = b$ is given by $\frac{1}{2}\left( {g^{2}(b) - g^{2}(a)} \right).$

107\.

Let $g$ be a positive, increasing, and differentiable function on the interval $\left\lbrack {a,b} \right\rbrack,$ and let $k$ be a positive real number. Show that the volume of the solid under the surface $z = g\prime(x)$ and above the region bounded by $y = g(x),y = g(x) + k,x = a,$ and $x = b$ is given by $k\left( {g(b) - g(a)} \right).$

108\.

Find the volume of the solid situated in the first octant and determined by the planes $z = 2,$ $z = 0,x + y = 1,x = 0,\ \text{and}\ y = 0.$

109.

Find the volume of the solid situated in the first octant and bounded by the planes $x + 2y = 1,$ $x = 0,y = 0,z = 4,\ \text{and}\ z = 0.$

110\.

Find the volume of the solid bounded by the planes $x + y = 1,x - y = 1,x = 0,z = 0,$ and $z = 10.$

111.

Find the volume of the solid bounded by the planes $x + y = 1,x - y = 1,x + y = -1,$ $x - y = -1,z = 1\ \text{and}\ z = 0.$

112\.

Let $S_{1}$ and $S_{2}$ be the solids situated in the first octant under the planes $x + y + z = 1$ and $x + y + 2z = 1,$ respectively, and let $S$ be the solid situated between $S_{1},S_{2},x = 0,\ \text{and}\ y = 0.$

1. Find the volume of the solid $S_{1}.$

2. Find the volume of the solid $S_{2}.$

3. Find the volume of the solid $S$ by subtracting the volumes of the solids $S_{1}\ \text{and}\ S_{2}.$

113.

Let $S_{1}\ \text{and}\ S_{2}$ be the solids situated in the first octant under the planes $2x + 2y + z = 2$ and $x + y + z = 1,$ respectively, and let $S$ be the solid situated between $S_{1},S_{2},x = 0,\ \text{and}\ y = 0.$

1. Find the volume of the solid $S_{1}.$

2. Find the volume of the solid $S_{2}.$

3. Find the volume of the solid $S$ by subtracting the volumes of the solids $S_{1}\ \text{and}\ S_{2}.$

114\.

Let $S_{1}\ \text{and}\ S_{2}$ be the solids situated in the first octant under the plane $x + y + z = 2$ and under the sphere $x^{2} + y^{2} + z^{2} = 4,$ respectively. If the volume of the solid $S_{2}$ is $\frac{4\pi}{3},$ determine the volume of the solid $S$ situated between $S_{1}$ and $S_{2}$ by subtracting the volumes of these solids.

115.

Consider the plane $x~ + ~y~ + ~z~ = ~2$ and the cylinder $x^{2}~ + ~y^{2}~ = ~~4~$in the first octant.

1. Find the volume under the plane.

2. Find the volume inside the cylinder under the plane $z~ = ~2$.

3. Find the volume above the plane, inside the cylinder, and below the plane $z~ = ~2$.

116\.

\[T\] The following figure shows the region $D$ bounded by the curves $y = \text{sin}\ x,$ $x = 0,$ and $y = x^{4}.$ Use a graphing calculator or CAS to find the $x$-coordinates of the intersection points of the curves and to determine the area of the region $D.$ Round your answers to six decimal places.

117.

\[T\] The region $D$ bounded by the curves $y = \text{cos}\ x,x = 0,\ \text{and}\ y = x^{3}$ is shown in the following figure. Use a graphing calculator or CAS to find the *x*-coordinates of the intersection points of the curves and to determine the area of the region $D.$ Round your answers to six decimal places.

118\.

Suppose that $\left( {X,Y} \right)$ is the outcome of an experiment that must occur in a particular region $S$ in the $xy$-plane. In this context, the region $S$ is called the sample space of the experiment and $X\ \text{and}\ Y$ are random variables. If $D$ is a region included in $S,$ then the probability of $\left( {X,Y} \right)$ being in $D$ is defined as $P\lbrack\left( {X,Y} \right) \in D\rbrack = {\iint\limits_{D}{p(x,y)dx\ dy}},$ where $p(x,y)$ is the joint probability density of the experiment. Here, $p(x,y)$ is a nonnegative function for which ${\iint\limits_{S}{p(x,y)dx\ dy = 1}}.$ Assume that a point $\left( {X,Y} \right)$ is chosen arbitrarily in the square $\lbrack 0,3\rbrack\ \times \ \lbrack 0,3\rbrack$ with the probability density

$p(x,y) = \begin{cases} \frac{1}{9} & {(x,y) \in \lbrack 0,3\rbrack\ \times \ \lbrack 0,3\rbrack,} \\ 0 & {\text{otherwise}\text{.}} \end{cases}$

Find the probability that the point $\left( {X,Y} \right)$ is inside the unit square and interpret the result.

119.

Consider $X\ \text{and}\ Y$ two random variables of probability densities $p_{1}(x)$ and $p_{2}(y),$ respectively. The random variables $X\ \text{and}\ Y$ are said to be independent if their joint density function is given by $p(x,y) = p_{1}(x)p_{2}(y).$ At a drive-thru restaurant, customers spend, on average, $3$ minutes placing their orders and an additional $5$ minutes paying for and picking up their meals. Assume that placing the order and paying for/picking up the meal are two independent events $X$ and $Y.$ If the waiting times are modeled by the exponential probability densities

$\begin{array}{lcclccl} {p_{1}(x) = \begin{cases} {\frac{1}{3}e^{\text{−}{x\text{/}3}}} & {x \geq 0,} \\ 0 & \text{otherwise,} \end{cases}} & & & \text{and} & & & {p_{2}(y) = \begin{cases} {\frac{1}{5}e^{\text{−}{y\text{/}5}}} & {y \geq 0,} \\ 0 & \text{otherwise,} \end{cases}} \end{array}$

respectively, the probability that a customer will spend less than 6 minutes in the drive-thru line is given by $P\left\lbrack {X + Y \leq 6} \right\rbrack = {\iint\limits_{D}{p(x,y)dx\ dy}},$ where $D = \left\{ \left. (x,y) \right\} \middle| x \geq 0,y \geq 0,x + y \leq 6 \right\}.$ Find $P\left\lbrack {X + Y \leq 6} \right\rbrack$ and interpret the result.

120\.

\[T\] The Reuleaux triangle consists of an equilateral triangle and three regions, each of them bounded by a side of the triangle and an arc of a circle of radius *s* centered at the opposite vertex of the triangle. Show that the area of the Reuleaux triangle in the following figure of side length $s$ is $\frac{s^{2}}{2}\left( {\pi - \sqrt{3}} \right).$

121\.

\[T\] Show that the area of the lunes of Alhazen, the two blue lunes in the following figure, is the same as the area of the right triangle *ABC*. The outer boundaries of the lunes are semicircles of diameters $AB\ \text{and}\ BC,$ respectively, and the inner boundaries are formed by the circumcircle of the triangle $ABC.$

---

5.3 Double Integrals in Polar Coordinates 5.3 极坐标下的二重积分

• 5.3.1 Recognize the format of a double integral over a polar rectangular region.
• 5.3.2 Evaluate a double integral in polar coordinates by using an iterated integral.
• 5.3.3 Recognize the format of a double integral over a general polar region.
• 5.3.4 Use double integrals in polar coordinates to calculate areas and volumes.

• 5.3.1 识别极矩形区域上二重积分的形式。
• 5.3.2 利用累次积分计算极坐标下的二重积分。
• 5.3.3 识别一般极坐标区域上二重积分的形式。
• 5.3.4 利用极坐标下的二重积分计算面积与体积。

Double integrals are sometimes much easier to evaluate if we change rectangular coordinates to polar coordinates. However, before we describe how to make this change, we need to establish the concept of a double integral in a polar rectangular region.

Analysis 分析

Note that if we were to find the volume of an arbitrary cone with radius $a$ units and height $h$ units, then the equation of the cone would be $z = h - \frac{h}{a}\sqrt{x^{2} + y^{2}}.$

We can still use Figure 5.37 and set up the integral as ${\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{r = 0}^{r = a}\left( {h - \frac{h}{a}r} \right)}}r\ dr\ d\theta.$

Evaluating the integral, we get $\frac{1}{3}\pi a^{2}h.$

Use polar coordinates to find an iterated integral for finding the volume of the solid enclosed by the paraboloids $z = x^{2} + y^{2}$ and $z = 16 - x^{2} - y^{2}.$

As with rectangular coordinates, we can also use polar coordinates to find areas of certain regions using a double integral. As before, we need to understand the region whose area we want to compute. Sketching a graph and identifying the region can be helpful to realize the limits of integration. Generally, the area formula in double integration will look like

$$\text{Area}\ A = {\int\limits_{\alpha}^{\beta}\ {\int\limits_{h_{1}{(\theta)}}^{h_{2}{(\theta)}}{1r\ dr\ d\theta}}}.$$

Finding an Area Using a Double Integral in Polar Coordinates 用极坐标下的二重积分求面积

Evaluate the area bounded by the curve $r = \text{cos}\ 4\theta.$

Solution 解

Sketching the graph of the function $r = \text{cos}\ 4\theta$ reveals that it is a polar rose with eight petals (see the following figure).

Using symmetry, we can see that we need to find the area of one petal and then multiply it by $8.$ Notice that the values of $\theta$ for which the graph passes through the origin are the zeros of the function $\text{cos}\ 4\theta,$ and these are odd multiples of $\pi\text{/}8.$ Thus, one of the petals corresponds to the values of $\theta$ in the interval $\left\lbrack {\text{−}\pi\text{/}8,\pi\text{/}8} \right\rbrack.$ Therefore, the area bounded by the curve $r = \text{cos}\ 4\theta$ is

$$\begin{array}{cl} A & {= 8{\int\limits_{\theta = \text{−}\pi\text{/}8}^{\theta = \pi\text{/}8}\ {\int\limits_{r = 0}^{r = \text{cos}\ 4\theta}{1r\ dr\ d\theta}}}} \\ & {= 8{\int\limits_{\text{−}\pi\text{/}8}^{\pi\text{/}8}\left\lbrack {\frac{1}{2}\left. r^{2} \right|_{0}^{\text{cos}\ 4\theta}} \right\rbrack}d\theta = 8{\int\limits_{\text{−}\pi\text{/}8}^{\pi\text{/}8}{\frac{1}{2}\text{cos}^{2}4\theta}}\ d\theta = 8\left\lbrack {\frac{1}{4}\theta + \left. {\frac{1}{16}\text{sin}\ 4\theta\ \text{cos}\ 4\theta} \right|_{\text{−}\pi\text{/}8}^{\pi\text{/}8}} \right\rbrack = 8\left\lbrack \frac{\pi}{16} \right\rbrack = \frac{\pi}{2}.} \end{array}$$

Finding Area Between Two Polar Curves 求两条极坐标曲线之间的面积

Find the area enclosed by the circle $r = 3\ \text{cos}\ \theta$ and the cardioid $r = 1 + \text{cos}\ \theta.$

Solution 解

First and foremost, sketch the graphs of the region (Figure 5.39).

We can from see the symmetry of the graph that we need to find the points of intersection. Setting the two equations equal to each other gives

$$3\ \text{cos}\ \theta = 1 + \text{cos}\ \theta.$$

One of the points of intersection is $\theta = \pi\text{/}3.$ The area above the polar axis consists of two parts, with one part defined by the cardioid from $\theta = 0$ to $\theta = \pi\text{/}3$ and the other part defined by the circle from $\theta = \pi\text{/}3$ to $\theta = \pi\text{/}2.$ By symmetry, the total area is twice the area above the polar axis. Thus, we have

$$A = 2\left\lbrack {\int\limits_{\theta = 0}^{\theta = \pi\text{/}3}\ {\int\limits_{r = 0}^{r = 1 + \text{cos}\ \theta}{1r\ dr\ d\theta + {\int\limits_{\theta = \pi\text{/}3}^{\theta = \pi\text{/}2}\ {\int\limits_{r = 0}^{r = 3\ \text{cos}\ \theta}{1r\ dr\ d\theta}}}}}} \right\rbrack.$$

Evaluating each piece separately, we find that the area is

$$A = 2\left( {\frac{1}{4}\pi + \frac{9}{16}\sqrt{3} + \frac{3}{8}\pi - \frac{9}{16}\sqrt{3}} \right) = 2\left( {\frac{5}{8}\pi} \right) = \frac{5}{4}\pi\ \text{square units}\text{.}$$

Find the area enclosed inside the cardioid $r = 3 - 3\ \text{sin}\ \theta$ and outside the cardioid $r = 1 + \text{sin}\ \theta.$

Evaluating an Improper Double Integral in Polar Coordinates 在极坐标下计算反常二重积分

Evaluate the integral ${\iint\limits_{\mathbb{R}^{2}}e^{-10{({x^{2} + y^{2}})}}}dx\ dy.$

Solution 解

This is an improper integral because we are integrating over an unbounded region $\mathbb{R}^{2}.$ In polar coordinates, the entire plane $\mathbb{R}^{2}$ can be seen as $0 \leq \theta \leq 2\pi,$ $0 \leq r < \infty.$

Using the changes of variables from rectangular coordinates to polar coordinates, we have

$$\begin{array}{cl} {{\iint\limits_{\mathbb{R}^{2}}e^{-10{({x^{2} + y^{2}})}}}dx\ dy} & {= {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{r = 0}^{r = \infty}{e^{-10r^{2}}r\ dr\ d\theta}}} = {\int\limits_{\theta = 0}^{\theta = 2\pi}\left( {\underset{a\rightarrow\infty}{\text{lim}}{\int\limits_{r = 0}^{r = a}{e^{-10r^{2}}r\ dr}}} \right)}d\theta} \\ & {= \left( {\int\limits_{\theta = 0}^{\theta = 2\pi}{d\theta}} \right)\left( {\underset{a\rightarrow\infty}{\text{lim}}{\int\limits_{r = 0}^{r = a}{e^{-10r^{2}}r\ dr}}} \right)} \\ & {= 2\pi\left( {\underset{a\rightarrow\infty}{\text{lim}}{\int\limits_{r = 0}^{r = a}{e^{-10r^{2}}r\ dr}}} \right)} \\ & {= 2\pi\underset{a\rightarrow\infty}{\text{lim}}\left( {- \frac{1}{20}} \right)\left( \left. e^{-10r^{2}} \right|_{0}^{a} \right)} \\ & {= 2\pi\left( {- \frac{1}{20}} \right)\underset{a\rightarrow\infty}{\text{lim}}\left( {e^{-10a^{2}} - 1} \right)} \\ & {= \frac{\pi}{10}.} \end{array}$$

Evaluate the integral ${\iint\limits_{\mathbb{R}^{2}}{e^{-4{({x^{2} + y^{2}})}}dx\ dy}}.$

Section 5.3 Exercises 5.3 节习题

In the following exercises, express the region $D$ in polar coordinates.

122\.

$D$ is the region of the disk of radius $2$ centered at the origin that lies in the first quadrant.

123.

$D$ is the region between the circles of radius $4$ and radius $5$ centered at the origin that lies in the second quadrant.

124\.

$D$ is the region bounded by the $y$-axis and $x = \sqrt{1 - y^{2}}.$

125.

$D$ is the region bounded by the $x$-axis and $y = \sqrt{2 - x^{2}}.$

126\.

$D = \left\{ {\left. \left( {x,y} \right) \right|x^{2} + y^{2} \leq 4x} \right\}$

127.

$D = \left\{ {\left. \left( {x,y} \right) \right|x^{2} + y^{2} \leq 4y} \right\}$

In the following exercises, the graph of the polar rectangular region $D$ is given. Express $D$ in polar coordinates.

128\. 129. 130. 131. 132.

In the following graph, the region $D$ is situated below $y = x$ and is bounded by $x = 1,x = 5,$ and $y = 0.$

133.

In the following graph, the region $D$ is bounded by $y = x$ and $y = x^{2}.$

In the following exercises, evaluate the double integral $\iint\limits_{R}{f\left( {x,y} \right)dA}$ over the polar rectangular region $D.$

134\.

$f\left( {x,y} \right) = x^{2} + y^{2},D = \left\{ {\left. \left( {r,\theta} \right) \right|3 \leq r \leq 5,0 \leq \theta \leq 2\pi} \right\}$

135.

$f\left( {x,y} \right) = x + y,\ \ D = \left\{ {\left. \left( {r,\theta} \right) \right|3 \leq r \leq 5,0 \leq \theta \leq 2\pi} \right\}$

136\.

$f\left( {x,y} \right) = x^{2} + xy,D = \left\{ {\left. \left( {r,\theta} \right) \right|1 \leq r \leq 2,\pi \leq \theta \leq 2\pi} \right\}$

137.

$f\left( {x,y} \right) = x^{4} + y^{4},D = \left\{ {\left. \left( {r,\theta} \right) \right|1 \leq r \leq 2,\frac{3\pi}{2} \leq \theta \leq 2\pi} \right\}$

138\.

$f\left( {x,y} \right) = \sqrt[3]{x^{2} + y^{2}},$ where $D = \left\{ {\left. \left( {r,\theta} \right) \right|0 \leq r \leq 1,\frac{\pi}{2} \leq \theta \leq \pi} \right\}.$

139.

$f\left( {x,y} \right) = x^{4} + 2x^{2}y^{2} + y^{4},$ where $D = \left\{ {\left. \left( {r,\theta} \right) \right|3 \leq r \leq 4,\frac{\pi}{3} \leq \theta \leq \frac{2\pi}{3}} \right\}.$

140\.

$f(x,y) = \text{sin}\left( {\text{arctan}\ \frac{y}{x}} \right),$ where $D = \left\{ (r,\theta) \middle| 1 \leq r \leq 2,\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3} \right\}$

141.

$f(x,y) = \text{arctan}\left( \frac{y}{x} \right),$ where $D = \left\{ (r,\theta) \middle| 2 \leq r \leq 3,\frac{\pi}{4} \leq \theta \leq \frac{\pi}{3} \right\}$

142\.

${\iint\limits_{D}{e^{x^{2} + y^{2}}\left\lbrack {1 + 2\ \text{arctan}\left( \frac{y}{x} \right)} \right\rbrack}}dA\text{,}\ D = \left\{ {\left. \left( {r,\theta} \right) \right|1 \leq r \leq 2,\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3}} \right\}$

143.

${\iint\limits_{D}\left( {e^{x^{2} + y^{2}} + x^{4} + 2x^{2}y^{2} + y^{4}} \right)}\text{arctan}\left( \frac{y}{x} \right)dA\text{,}\ D = \left\{ {\left. \left( {r,\theta} \right) \right|1 \leq r \leq 2,\frac{\pi}{4} \leq \theta \leq \frac{\pi}{3}} \right\}$

In the following exercises, the integrals have been converted to polar coordinates. Verify that the identities are true and choose the easiest way to evaluate the integrals, in rectangular or polar coordinates.

144\.

${\int\limits_{1}^{2}\ {\int\limits_{0}^{x}\left( {x^{2} + y^{2}} \right)}}dy\ dx = {\int\limits_{0}^{\frac{\pi}{4}}\ {\int\limits_{\text{sec}\ \theta}^{2\ \text{sec}\ \theta}{r^{3}dr\ d\theta}}}$

145.

${\int\limits_{2}^{3}\ {\int\limits_{0}^{x}{\frac{x}{\sqrt{x^{2} + y^{2}}}dy\ dx}}} = {\int\limits_{0}^{\pi\text{/}4}\ {\int\limits_{2\ \text{sec}\ \theta}^{3\text{sec}\ \theta}{r\ \text{cos}\ \theta\ dr\ d\theta}}}$

146\.

${\int\limits_{0}^{1}\ {\int\limits_{x^{2}}^{x}{\frac{1}{\sqrt{x^{2} + y^{2}}}dy\ dx}}} = {\int\limits_{0}^{\pi\text{/}4}\ {\int\limits_{0}^{\text{tan}\ \theta\ \text{sec}\ \theta}{dr\ d\theta}}}$

147.

${\int\limits_{0}^{1}\ {\int\limits_{x^{2}}^{x}{\frac{y}{\sqrt{x^{2} + y^{2}}}dy\ dx}}} = {\int\limits_{0}^{\pi\text{/}4}\ {\int\limits_{0}^{\text{tan}\ \theta\ \text{sec}\ \theta}{r\ \text{sin}\ \theta\ dr\ d\theta}}}$

In the following exercises, convert the integrals to polar coordinates and evaluate them.

148\.

$\int\limits_{0}^{3}\ {\int\limits_{0}^{\sqrt{9 - y^{2}}}{\left( {x^{2} + y^{2}} \right)dx\ dy}}$

149.

$\int\limits_{0}^{2}\ {\int\limits_{\text{−}\sqrt{4 - y^{2}}}^{\sqrt{4 - y^{2}}}{\left( {x^{2} + y^{2}} \right)^{2}dx\ dy}}$

150\.

$\int\limits_{0}^{1}\ {\int\limits_{0}^{\sqrt{1 - x^{2}}}{\left( {x + y} \right)dy\ dx}}$

151.

$\int\limits_{0}^{4}\ {\int\limits_{\text{−}\sqrt{16 - x^{2}}}^{\sqrt{16 - x^{2}}}{\text{sin}\left( {x^{2} + y^{2}} \right)dy\ dx}}$

152\.

Evaluate the integral $\iint\limits_{D}{r\ dA}$ where $D$ is the region bounded by the polar axis and the upper half of the cardioid $r = 1 + \text{cos}\ \theta.$

153.

Find the area of the region $D$ bounded by the polar axis and the upper half of the cardioid $r = 1 + \text{cos}\ \theta.$

154\.

Evaluate the integral ${\iint\limits_{D}{\ dA}},$ where $D$ is the region bounded by the part of the four-leaved rose $r = \text{sin}\ 2\theta$ situated in the first quadrant (see the following figure).

155.

Find the total area of the region enclosed by the four-leaved rose $r = \text{sin}\ 2\theta$ (see the figure in the previous exercise).

156\.

Find the area of the region $D,$ which is the region bounded by $y = \sqrt{4 - x^{2}},$ $x = \sqrt{3},$ $x = 2,$ and $y = 0.$

157.

Find the area of the region $D,$ which is the region inside the disk $x^{2} + y^{2} \leq 4$ and to the right of the line $x = 1.$

158\.

Determine the average value of the function $f\left( {x,y} \right) = x^{2} + y^{2}$ over the region $D$ bounded by the polar curve $r = \text{cos}\ 2\theta,$ where $- \frac{\pi}{4} \leq \theta \leq \frac{\pi}{4}$ (see the following graph).

159.

Determine the average value of the function $f\left( {x,y} \right) = \sqrt{x^{2} + y^{2}}$ over the region $D$ bounded by the polar curve $r = 3\ \text{sin}\ 2\theta,$ where $0 \leq \theta \leq \frac{\pi}{2}$ (see the following graph).

160\.

Find the volume of the solid situated in the first octant and bounded by the paraboloid $z = 1 - 4x^{2} - 4y^{2}$ and the planes $x = 0,y = 0,$ and $z = 0.$

161.

Find the volume of the solid bounded by the paraboloid $z = 2 - 9x^{2} - 9y^{2}$ and the plane $z = 1.$

162\.

1. Find the volume of the solid $S_{1}$ bounded by the cylinder $x^{2} + y^{2} = 1$ and the planes $z = 0$ and $z = 1.$

2. Find the volume of the solid $S_{2}$ outside the double cone $z^{2} = x^{2} + y^{2},$ inside the cylinder $x^{2} + y^{2} = 1,$ and above the plane $z = 0.$

3. Find the volume of the solid inside the cone $z^{2} = x^{2} + y^{2}$ and below the plane $z = 1$ by subtracting the volumes of the solids $S_{1}$ and $S_{2}.$

163.

1. Find the volume of the solid $S_{1}$ inside the unit sphere $x^{2} + y^{2} + z^{2} = 1$ and above the plane $z = 0.$

2. Find the volume of the solid $S_{2}$ inside the double cone $\left( {z - 1} \right)^{2} = x^{2} + y^{2}$ and above the plane $z = 0.$

3. Find the volume of the solid outside the double cone $\left( {z - 1} \right)^{2} = x^{2} + y^{2}$ and inside the sphere $x^{2} + y^{2} + z^{2} = 1.$

For the following two exercises, consider a spherical ring, which is a sphere with a cylindrical hole cut so that the axis of the cylinder passes through the center of the sphere (see the following figure).

164\.

If the sphere has radius $4$ and the cylinder has radius $2,$ find the volume of the spherical ring.

165.

A cylindrical hole of diameter $6$ cm is bored through a sphere of radius $5$ cm such that the axis of the cylinder passes through the center of the sphere. Find the volume of the resulting spherical ring.

166\.

Find the volume of the solid that lies under the double cone $z^{2} = 4x^{2} + 4y^{2},$ inside the cylinder $x^{2} + y^{2} = x,$ and above the plane $z = 0.$

167.

Find the volume of the solid that lies under the paraboloid $z = x^{2} + y^{2},$ inside the cylinder $x^{2} + y^{2} = x,$ and above the plane $z = 0.$

168\.

Find the volume of the solid that lies under the plane $x + y + z = 10$ and above the disk $x^{2} + y^{2} = 4x.$

169.

Find the volume of the solid that lies under the plane $2x + y + 2z = 8$ and above the unit disk $x^{2} + y^{2} = 1.$

170\.

A radial function $f$ is a function whose value at each point depends only on the distance between that point and the origin of the system of coordinates; that is, $f(x,y) = g(r),$ where $r = \sqrt{x^{2} + y^{2}}.$ Show that if $f$ is a continuous radial function, then ${\iint\limits_{D}{f\left( {x,y} \right)}}dA = \left( {\theta_{2} - \theta_{1}} \right)\left\lbrack {G(R_{2}) - G(R_{1})} \right\rbrack,$ where $G\prime(r) = rg(r)$ and $\left( {x,y} \right) \in D = \left\{ (r,\theta) \middle| R_{1} \leq r \leq R_{2},\theta_{1} \leq \theta \leq \theta_{2} \right\},$ with $0 \leq R_{1} < R_{2}$ and $0 \leq \theta_{1} < \theta_{2} \leq 2\pi.$

171.

Use the information from the preceding exercise to calculate the integral ${\iint\limits_{D}\left( {x^{2} + y^{2}} \right)^{3}}dA,$ where $D$ is the unit disk.

172\.

Let $f(x,y) = \frac{F\prime(r)}{r}$ be a continuous radial function defined on the annular region $D = \left\{ (r,\theta) \middle| R_{1} \leq r \leq R_{2},0 \leq \theta \leq 2\pi \right\},$ where $r = \sqrt{x^{2} + y^{2}},$ $0 < R_{1} < R_{2},$ and $F$ is a differentiable function. Show that ${\iint\limits_{D}{f\left( {x,y} \right)}}dA = 2\pi\left\lbrack {F(R_{2}) - F(R_{1})} \right\rbrack.$

173.

Apply the preceding exercise to calculate the integral ${\iint\limits_{D}\frac{e^{\sqrt{x^{2} + y^{2}}}}{\sqrt{x^{2} + y^{2}}}}dx\ dy,$ where $D$ is the annular region between the circles of radii $1$ and $2$ situated in the third quadrant.

174\.

Let $f$ be a continuous function that can be expressed in polar coordinates as a product of a function of $r$ only and a function of $\theta$ only; that is, $f(x,y) = h(\theta),$ where $\left( {x,y} \right) \in D = \left\{ (r,\theta) \middle| R_{1} \leq r \leq R_{2},\theta_{1} \leq \theta \leq \theta_{2} \right\},$ with $0 \leq R_{1} < R_{2}$ and $0 \leq \theta_{1} < \theta_{2} \leq 2\pi.$ Show that ${\iint\limits_{D}{f\left( {x,y} \right)}}dA = \frac{1}{2}\left( {R_{2}^{2} - R_{1}^{2}} \right)\left\lbrack {H(\theta_{2}) - H(\theta_{1})} \right\rbrack,$ where $H$ is an antiderivative of $h.$

175.

Apply the preceding exercise to calculate the integral ${\iint\limits_{D}\frac{y^{2}}{x^{2}}}dA,$ where $D = \left\{ (r,\theta) \middle| 1 \leq r \leq 2,\frac{\pi}{6} \leq \theta \leq \frac{\pi}{3} \right\}.$

176\.

Let $f$ be a continuous function that can be expressed in polar coordinates as a product of a function of $r$ only and function of $\theta$ only; that is, $f(x,y) = g(r)h(\theta),$ where $\left( {x,y} \right) \in D = \left\{ (r,\theta) \middle| R_{1} \leq r \leq R_{2},\theta_{1} \leq \theta \leq \theta_{2} \right\}$ with $0 \leq R_{1} < R_{2}$ and $0 \leq \theta_{1} < \theta_{2} \leq 2\pi.$ Show that ${\iint\limits_{D}{f\left( {x,y} \right)}}dA = \left\lbrack {G(R_{2}) - G(R_{1})} \right\rbrack\ \left\lbrack {H(\theta_{2}) - H(\theta_{1})} \right\rbrack,$ where $G(r)$ and $H(\theta)$ are antiderivatives of $rg(r)$ and $h(\theta)$, respectively.

177.

Evaluate ${\iint\limits_{D}{\text{arctan}\left( \frac{y}{x} \right)\sqrt{x^{2} + y^{2}}}}dA,$ where $D = \left\{ (r,\theta) \middle| 2 \leq r \leq 3,\frac{\pi}{4} \leq \theta \leq \frac{\pi}{3} \right\}.$

178\.

A spherical cap is the region of a sphere that lies above or below a given plane.

1. Show that the volume of the spherical cap in the figure below is $\frac{1}{6}\pi h\left( {3a^{2} + h^{2}} \right).$

2. A spherical segment is the solid defined by intersecting a sphere with two parallel planes. If the distance between the planes is $h,$ show that the volume of the spherical segment in the figure below is $\frac{1}{6}\pi h\left( {3a^{2} + 3b^{2} + h^{2}} \right).$

179.

In statistics, the joint density for two independent, normally distributed events with a mean $\mu = 0$ and a standard distribution $\sigma$ is defined by $p(x,y) = \frac{1}{2\pi\sigma^{2}}e^{- \frac{x^{2} + y^{2}}{2\sigma^{2}}}.$ Consider $\left( {X,Y} \right),$ the Cartesian coordinates of a ball in the resting position after it was released from a position on the *z*-axis toward the $xy$-plane. Assume that the coordinates of the ball are independently normally distributed with a mean $\mu = 0$ and a standard deviation of $\sigma$ (in feet). The probability that the ball will stop no more than $a$ feet from the origin is given by $P\left\lbrack {X^{2} + Y^{2} \leq a^{2}} \right\rbrack = {\iint\limits_{D}{p(x,y)dy\ dx}},$ where $D$ is the disk of radius *a* centered at the origin. Show that $P\left\lbrack {X^{2} + Y^{2} \leq a^{2}} \right\rbrack = 1 - e^{\text{−}{a^{2}\text{/}{2\sigma^{2}}}}.$

180\.

The double improper integral ${\int\limits_{\text{−}\infty}^{\infty}\ {\int\limits_{\text{−}\infty}^{\infty}e^{\text{−}({{x^{2} + y^{2}})\text{/}2}}}}dy\ dx$ may be defined as the limit value of the double integrals ${\iint\limits_{D_{a}}e^{\text{−}({x^{2} + y^{2}})\text{/}2}}dA$ over disks $D_{a}$ of radii *a* centered at the origin, as *a* increases without bound; that is, ${\int\limits_{\text{−}\infty}^{\infty}\ {\int\limits_{\text{−}\infty}^{\infty}e^{\text{−}({x^{2} + y^{2}})\text{/}2}}}dy\ dx = \underset{a\rightarrow\infty}{\text{lim}}{\iint\limits_{D_{a}}e^{\text{−}({x^{2} + y^{2}})\text{/}2}}dA.$

1. Use polar coordinates to show that ${\int\limits_{\text{−}\infty}^{\infty}\ {\int\limits_{\text{−}\infty}^{\infty}e^{\text{−}({x^{2} + y^{2}})\text{/}2}}}dy\ dx = 2\pi.$

2. Show that ${\int\limits_{\text{−}\infty}^{\infty}e^{\text{−}{x^{2}\text{/}2}}}dx = \sqrt{2\pi},$ by using the relation ${\int\limits_{\text{−}\infty}^{\infty}\ {\int\limits_{\text{−}\infty}^{\infty}e^{\text{−}({x^{2} + y^{2}})\text{/}2}}}dy\ dx = \left( {\int\limits_{\text{−}\infty}^{\infty}{e^{\text{−}{x^{2}\text{/}2}}dx}} \right)\left( {{\int\limits_{\text{−}\infty}^{\infty}e^{\text{−}{y^{2}\text{/}2}}}dy} \right).$

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5.4 Triple Integrals 5.4 三重积分

• 5.4.1 Recognize when a function of three variables is integrable over a rectangular box.
• 5.4.2 Evaluate a triple integral by expressing it as an iterated integral.
• 5.4.3 Recognize when a function of three variables is integrable over a closed and bounded region.
• 5.4.4 Simplify a calculation by changing the order of integration of a triple integral.
• 5.4.5 Calculate the average value of a function of three variables.

• 5.4.1 识别三元函数在矩形盒上何时可积。
• 5.4.2 通过把三重积分写成累次积分来计算它。
• 5.4.3 识别三元函数在闭有界区域上何时可积。
• 5.4.4 通过交换三重积分的积分次序来简化计算。
• 5.4.5 计算三元函数的平均值。

In Double Integrals over Rectangular Regions, we discussed the double integral of a function $f(x,y)$ of two variables over a rectangular region in the plane. In this section we define the triple integral of a function $f(x,y,z)$ of three variables over a rectangular solid box in space, $\mathbb{R}^{3}.$ Later in this section we extend the definition to more general regions in $\mathbb{R}^{3}.$

Integrable Functions of Three Variables 三元函数的可积性

We can define a rectangular box $B$ in $\mathbb{R}^{3}$ as $B = \left\{ (x,y,z) \middle| a \leq x \leq b,c \leq y \leq d,e \leq z \leq f \right\}.$ We follow a similar procedure to what we did in Double Integrals over Rectangular Regions. We divide the interval $\lbrack a,b\rbrack$ into $l$ subintervals $\lbrack x_{i - 1},x_{i}\rbrack$ of equal length $\text{Δ}x = \frac{b - a}{l},$ divide the interval $\lbrack c,d\rbrack$ into $m$ subintervals $\lbrack y_{j - 1},y_{j}\rbrack$ of equal length $\text{Δ}y = \frac{d - c}{m},$ and divide the interval $\lbrack e,f\rbrack$ into $n$ subintervals $\lbrack z_{k - 1},z_{k}\rbrack$ of equal length $\text{Δ}z = \frac{f - e}{n}.$ Then the rectangular box $B$ is subdivided into $lmn$ subboxes $B_{ijk} = \lbrack x_{i - 1},x_{i}\rbrack\ \times \ \lbrack y_{j - 1},y_{j}\rbrack\ \times \ \lbrack z_{k - 1},z_{k}\rbrack,$ as shown in Figure 5.40.

For each $i,j,\ \text{and}\ k,$ consider a sample point $(x_{ijk}^{*},y_{ijk}^{*},z_{ijk}^{*})$ in each sub-box $B_{ijk}.$ We see that its volume is $\text{Δ}V = \text{Δ}x\text{Δ}y\text{Δ}z.$ Form the triple Riemann sum

$${\sum\limits_{i = 1}^{l}{\sum\limits_{j = 1}^{m}{\sum\limits_{k = 1}^{n}{f(x_{ijk}^{*},y_{ijk}^{*},z_{ijk}^{*})\text{Δ}x\text{Δ}y\text{Δ}z}}}}.$$

We define the triple integral in terms of the limit of a triple Riemann sum, as we did for the double integral in terms of a double Riemann sum.

The triple integral of a function $f(x,y,z)$ over a rectangular box $B$ is defined as

$$\underset{l,m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{l}{\sum\limits_{j = 1}^{m}{\sum\limits_{k = 1}^{n}{f(x_{ijk}^{*},y_{ijk}^{*},z_{ijk}^{*})\text{Δ}x\text{Δ}y\text{Δ}z = {\iiint\limits_{B}{f\left( {x,y,z} \right)dV}}}}}}$$ (5.10)

if this limit exists.

When the triple integral exists on $B,$ the function $f(x,y,z)$ is said to be integrable on $B.$ Also, the triple integral exists if $f(x,y,z)$ is continuous on $B.$ Therefore, we will use continuous functions for our examples. However, continuity is sufficient but not necessary; in other words, $f$ is bounded on $B$ and continuous except possibly on the boundary of $B.$ The sample point $(x_{ijk}^{*},y_{ijk}^{*},z_{ijk}^{*})$ can be any point in the rectangular sub-box $B_{ijk}$ and all the properties of a double integral apply to a triple integral. Just as the double integral has many practical applications, the triple integral also has many applications, which we discuss in later sections.

Now that we have developed the concept of the triple integral, we need to know how to compute it. Just as in the case of the double integral, we can have an iterated triple integral, and consequently, a version of Fubini’s thereom for triple integrals exists.

Fubini’s Theorem for Triple Integrals 三重积分的富比尼定理

If $f(x,y,z)$ is continuous on a rectangular box $B = \lbrack a,b\rbrack\ \times \ \lbrack c,d\rbrack\ \times \ \lbrack r,s\rbrack,$ then

$${\iiint\limits_{B}{f(x,y,z)dV}} = {\int\limits_{r}^{s}\ {\int\limits_{c}^{d}\ {\int\limits_{a}^{b}{f(x,y,z)dx\ dy\ dz}}}}.$$

This integral is also equal to any of the other five possible orderings for the iterated triple integral.

For $a,b,c,d,e,$ and $f$ real numbers, the iterated triple integral can be expressed in six different orderings:

$$\begin{array}{cl} {\int\limits_{r}^{s}\ {\int\limits_{c}^{d}\ {\int\limits_{a}^{b}{f(x,y,z)dx\ dy\ dz}}}} & {= {\int\limits_{r}^{s}{({\int\limits_{c}^{d}{({\int\limits_{a}^{b}{f(x,y,z)dx)dy)dz}}}}}} = {\int\limits_{c}^{d}{({\int\limits_{r}^{s}{({\int\limits_{a}^{b}{f(x,y,z)dx)dz)dy}}}}}}} \\ & {= {\int\limits_{a}^{b}{({\int\limits_{r}^{s}{({\int\limits_{c}^{d}{f(x,y,z)dy)dz)dx}}}}}} = {\int\limits_{r}^{s}{({\int\limits_{a}^{b}{({\int\limits_{c}^{d}{f(x,y,z)dy)dx)dz}}}}}}} \\ & {= {\int\limits_{c}^{e}{({\int\limits_{a}^{b}{({\int\limits_{r}^{s}{f(x,y,z)dz)dx)dy}}}}}} = {\int\limits_{a}^{b}{({\int\limits_{c}^{e}{({\int\limits_{r}^{s}{f(x,y,z)dz)dy)dx}}}}}}.} \end{array}$$

For a rectangular box, the order of integration does not make any significant difference in the level of difficulty in computation. We compute triple integrals using Fubini’s Theorem rather than using the Riemann sum definition. We follow the order of integration in the same way as we did for double integrals (that is, from inside to outside).

Evaluating a Triple Integral 计算三重积分

Evaluate the triple integral ${\int_{z = 0}^{z = 1}{\int_{y = 2}^{y = 4}{\int_{x = -1}^{x = 5}{(x + yz^{2})dx\ dy\ dz}}}}.$

Solution 解

The order of integration is specified in the problem, so integrate with respect to $x$ first, then *y*, and then $z.$

$$\begin{array}{lccc} & & & \\ & & & \\ & & & \\ & & & \\ {\quad{\int_{z = 0}^{z = 1}{\int_{y = 2}^{y = 4}{\int_{x = -1}^{x = 5}{(x + yz^{2})dx\ dy\ dz}}}}} & & & \\ {= {\int_{z = 0}^{z = 1}{\int_{y = 2}^{y = 4}{\left\lbrack \left. {\frac{x^{2}}{2} + xyz^{2}} \right|_{x = -1}^{x = 5} \right\rbrack dy\ dz}}}} & & & {\text{Integrate with respect to}\ x.} \\ {= {\int_{z = 0}^{z = 1}{\int_{y = 2}^{y = 4}{\left\lbrack {12 + 6yz^{2}} \right\rbrack dy\ dz}}}} & & & \text{Evaluate.} \\ {= {\int_{z = 0}^{z = 1}{\left\lbrack \left. {12y + 6\frac{y^{2}}{2}z^{2}} \right|_{y = 2}^{y = 4} \right\rbrack dz}}} & & & {\text{Integrate with respect to}\ y.} \\ {= {\int_{z = 0}^{z = 1}{\left\lbrack {24 + 36z^{2}} \right\rbrack dz}}} & & & \text{Evaluate.} \\ {= \left\lbrack {24z + 36\frac{z^{3}}{3}} \right\rbrack_{z = 0}^{z = 1} = 36.} & & & {\text{Integrate with respect to}\ z.} \end{array}$$

Evaluating a Triple Integral 计算三重积分

Evaluate the triple integral ${\iiint\limits_{B}{x^{2}yz}}\ dV$ where $B = \left\{ (x,y,z) \middle| - 2 \leq x \leq 1,0 \leq y \leq 3,1 \leq z \leq 5 \right\}$ as shown in the following figure.

Solution 解

The order is not specified, but we can use the iterated integral in any order without changing the level of difficulty. Choose, say, to integrate *y* first, then *x*, and then *z*.

$\begin{array}{cl} {{\iiint\limits_{B}{x^{2}yz}}\ dV} & {= {\int\limits_{1}^{5}\ {\int\limits_{-2}^{1}\ {\int\limits_{0}^{3}\left\lbrack {x^{2}yz} \right\rbrack}}}dy\ dx\ dz = {\int\limits_{1}^{5}\ {\int\limits_{-2}^{1}\left\lbrack \left. {x^{2}\frac{y^{2}}{2}z} \right|_{0}^{3} \right\rbrack}}dx\ dz} \\ & {= {\int\limits_{1}^{5}\ {\int\limits_{-2}^{1}\frac{9}{2}}}x^{2}z\ dx\ dz = {\int\limits_{1}^{5}\left\lbrack \left. {\frac{9}{2}\frac{x^{3}}{3}z} \right|_{-2}^{1} \right\rbrack}dz = {\int\limits_{1}^{5}{\frac{27}{2}z\ dz}} = \left. {\frac{27}{2}\frac{z^{2}}{2}} \right|_{1}^{5} = 162.} \end{array}$

Solution 解

The projection of the solid region $E$ onto the $xy$-plane is the region bounded above by $y = 4$ and below by the parabola $y = x^{2}$ as shown.

Thus, we have

$$E = \left\{ {\left. \left( {x,y,z} \right) \right| - 2 \leq x \leq 2,x^{2} \leq y \leq 4,\text{−}\sqrt{y - x^{2}} \leq z \leq \sqrt{y - x^{2}}} \right\}.$$

The triple integral becomes

$${\iiint\limits_{E}\sqrt{x^{2} + z^{2}}}dV = {\int\limits_{x = -2}^{x = 2}\ {\int\limits_{y = x^{2}}^{y = 4}\ {\int\limits_{z = \text{−}\sqrt{y - x^{2}}}^{z = \sqrt{y - x^{2}}}\sqrt{x^{2} + z^{2}}}}}dz\ dy\ dx.$$

This expression is difficult to compute, so consider the projection of $E$ onto the $xz$-plane. This is a circular disc $x^{2} + z^{2} \leq 4.$ So we obtain

$${\iiint\limits_{E}\sqrt{x^{2} + z^{2}}}dV = {\int\limits_{x = -2}^{x = 2}\ {\int\limits_{y = x^{2}}^{y = 4}\ {\int\limits_{z = \text{−}\sqrt{y - x^{2}}}^{z = \sqrt{y - x^{2}}}\sqrt{x^{2} + z^{2}}}}}dz\ dy\ dx = {\int\limits_{x = -2}^{x = 2}\ {\int\limits_{z = \text{−}\sqrt{4 - x^{2}}}^{z = \sqrt{4 - x^{2}}}\ {\int\limits_{y = x^{2} + z^{2}}^{y = 4}\sqrt{x^{2} + z^{2}}}}}dy\ dz\ dx.$$

Here the order of integration changes from being first with respect to $z,$ then $y,$ and then $x$ to being first with respect to $y,$ then to $z,$ and then to $x.$ It will soon be clear how this change can be beneficial for computation. We have

$${\int\limits_{x = -2}^{x = 2}\ {\int\limits_{z = \text{−}\sqrt{4 - x^{2}}}^{z = \sqrt{4 - x^{2}}}\ {\int\limits_{y = x^{2} + z^{2}}^{y = 4}\sqrt{x^{2} + z^{2}}}}}dy\ dz\ dx = {\int\limits_{x = -2}^{x = 2}\ {\int\limits_{z = \text{−}\sqrt{4 - x^{2}}}^{z = \sqrt{4 - x^{2}}}{\left( {4 - x^{2} - z^{2}} \right)\sqrt{x^{2} + z^{2}}}}}dz\ dx.$$

Now use the polar substitution $x = r\ \text{cos}\ \theta,z = r\ \text{sin}\ \theta,$ and $dz\ dx = r\ dr\ d\theta$ in the $xz$-plane. This is essentially the same thing as when we used polar coordinates in the $xy$-plane, except we are replacing $y$ by $z.$ Consequently the limits of integration change and we have, by using $r^{2} = x^{2} + z^{2},$

$$\begin{array}{cl} {{\int\limits_{x = -2}^{x = 2}\ {\int\limits_{z = \text{−}\sqrt{4 - x^{2}}}^{z = \sqrt{4 - x^{2}}}{\left( {4 - x^{2} - z^{2}} \right)\sqrt{x^{2} + z^{2}}}}}dz\ dx} & {= {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{r = 0}^{r = 2}\left( {4 - r^{2}} \right)}}rr\ dr\ d\theta} \\ & {= {\int\limits_{0}^{2\pi}\left\lbrack \left. {\frac{4r^{3}}{3} - \frac{r^{5}}{5}} \right|_{0}^{2} \right\rbrack}d\theta = {\int\limits_{0}^{2\pi}\frac{64}{15}}d\theta = \frac{128\pi}{15}.} \end{array}$$

Average Value of a Function of Three Variables 三元函数的平均值

Recall that we found the average value of a function of two variables by evaluating the double integral over a region on the plane and then dividing by the area of the region. Similarly, we can find the average value of a function in three variables by evaluating the triple integral over a solid region and then dividing by the volume of the solid.

Average Value of a Function of Three Variables 三元函数的平均值

If $f\left( {x,y,z} \right)$ is integrable over a solid bounded region $E$ with positive volume $V(E),$ then the average value of the function is

$$f_{\text{ave}} = \frac{1}{V(E)}{\iiint\limits_{E}{f\left( {x,y,z} \right)}}dV.$$

Note that the volume is $V(E) = {\iiint\limits_{E}{1dV}}.$

Finding an Average Temperature 求平均温度

The temperature at a point $\left( {x,y,z} \right)$ of a solid $E$ bounded by the coordinate planes and the plane $x + y + z = 1$ is $T(x,y,z) = (xy + 8z + 20)\text{°}\text{C}\text{.}$ Find the average temperature over the solid.

Solution 解

Use the theorem given above and the triple integral to find the numerator and the denominator. Then do the division. Notice that the plane $x + y + z = 1$ has intercepts $\left( {1,0,0} \right),\left( {0,1,0} \right),$ and $\left( {0,0,1} \right).$ The region $E$ looks like

$$E = \left\{ {\left. \left( {x,y,z} \right) \right|0 \leq x \leq 1,0 \leq y \leq 1 - x,0 \leq z \leq 1 - x - y} \right\}.$$

Hence the triple integral of the temperature is

$${\iiint\limits_{E}{f\left( {x,y,z} \right)}}dV = {\int\limits_{x = 0}^{x = 1}\ {\int\limits_{y = 0}^{y = 1 - x}\ {\int\limits_{z = 0}^{z = 1 - x - y}{\left( {xy + 8z + 20} \right)dz\ dy\ dx = \frac{147}{40}}}}}.$$

The volume evaluation is $V(E) = {\iiint\limits_{E}1}dV = {\int\limits_{x = 0}^{x = 1}\ {\int\limits_{y = 0}^{y = 1 - x}\ {\int\limits_{z = 0}^{z = 1 - x - y}{1dz\ dy\ dx = \frac{1}{6}}}}}.$

Hence the average value is $T_{\text{ave}} = \frac{147\text{/}40}{1\text{/}6} = \frac{6(147)}{40} = \frac{441}{20}$ degrees Celsius.

Find the average value of the function $f\left( {x,y,z} \right) = xyz$ over the cube with sides of length $4$ units in the first octant with one vertex at the origin and edges parallel to the coordinate axes.

Section 5.4 Exercises 5.4 节习题

In the following exercises, evaluate the triple integrals over the rectangular solid box $B.$

181.

${\iiint\limits_{B}\left( {2x + 3y^{2} + 4z^{3}} \right)}dV,$ where $B = \left\{ {\left. \left( {x,y,z} \right) \right|0 \leq x \leq 1,0 \leq y \leq 2,0 \leq z \leq 3} \right\}$

182\.

${\iiint\limits_{B}\left( {xy + yz + xz} \right)}dV,$ where $B = \left\{ {\left. \left( {x,y,z} \right) \right|1 \leq x \leq 2,0 \leq y \leq 2,1 \leq z \leq 3} \right\}$

183.

${\iiint\limits_{B}\left( {x\ \text{cos}\ y + z} \right)}dV,$ where $B = \left\{ {\left. \left( {x,y,z} \right) \right|0 \leq x \leq 1,0 \leq y \leq \pi,-1 \leq z \leq 1} \right\}$

184\.

${\iiint\limits_{B}\left( {z\ \text{sin}\ x + y^{2}} \right)}dV,$ where $B = \left\{ {\left. \left( {x,y,z} \right) \right|0 \leq x \leq \pi,0 \leq y \leq 1,-1 \leq z \leq 2} \right\}$

In the following exercises, change the order of integration by integrating first with respect to $z,$ then $x,$ then $y.$

185.

${\int\limits_{0}^{1}\ {\int\limits_{1}^{2}\ {\int\limits_{2}^{3}\left( {x^{2} + \text{ln}\ y + z} \right)}}}dx\ dy\ dz$

186\.

$\int\limits_{- 1}^{1}\ \int\limits_{0}^{3}\ \int\limits_{0}^{1}\left( ze^{x} + 2y \right)dx\ dy\ dz$

187.

${\int\limits_{-1}^{2}\ {\int\limits_{1}^{3}\ {\int\limits_{0}^{4}\left( {x^{2}z + \frac{1}{y}} \right)}}}dx\ dy\ dz$

188\.

${\int\limits_{1}^{2}\ {\int\limits_{-2}^{-1}\ {\int\limits_{0}^{1}\frac{x + y}{z}}}}dx\ dy\ dz$

189\.

Let $F,G,\ \text{and}\ H$ be continuous functions on $\left\lbrack {a,b} \right\rbrack,\left\lbrack {c,d} \right\rbrack,$ and $\left\lbrack {e,f} \right\rbrack,$ respectively, where $a,b,c,d,e,\ \text{and}\ f$ are real numbers such that $a < b,c < d,\ \text{and}\ e < f.$ Show that

$${\int\limits_{a}^{b}\ {\int\limits_{c}^{d}\ {\int\limits_{e}^{f}{F(x)}}}}G(y)H(z)dz\ dy\ dx = \left( {\int\limits_{a}^{b}{F(x)dx}} \right)\left( {\int\limits_{c}^{d}{G(y)dy}} \right)\left( {\int\limits_{e}^{f}{H(z)dz}} \right).$$ 190.

Let $F,G,\ \text{and}\ H$ be differential functions on $\left\lbrack {a,b} \right\rbrack,\left\lbrack {c,d} \right\rbrack,$ and $\left\lbrack {e,f} \right\rbrack,$ respectively, where $a,b,c,d,e,\ \text{and}\ f$ are real numbers such that $a < b,c < d,\ \text{and}\ e < f.$ Show that

$${\int\limits_{a}^{b}\ {\int\limits_{c}^{d}\ {\int\limits_{e}^{f}{F^{\prime}(x)}}}}G^{\prime}(y)H^{\prime}(z)dz\ dy\ dx = \left\lbrack {F(b) - F(a)} \right\rbrack\ \left\lbrack {G(d) - G(c)} \right\rbrack\ \left\lbrack {H(f) - H(e)} \right\rbrack.$$

In the following exercises, evaluate the triple integrals over the bounded region $E = \left\{ {\left. \left( {x,y,z} \right) \right|a \leq x \leq b,h_{1}(x) \leq y \leq h_{2}(x),e \leq z \leq f} \right\}.$

191.

${\iiint\limits_{E}\left( {2x + 5y + 7z} \right)}dV,$ where $E = \left\{ {\left. \left( {x,y,z} \right) \right|0 \leq x \leq 1,0 \leq y \leq - x + 1,1 \leq z \leq 2} \right\}$

192\.

${\iiint\limits_{E}\left( {y\ \text{ln}\ x + z} \right)}dV,$ where $E = \left\{ {\left. \left( {x,y,z} \right) \right|1 \leq x \leq e,0 \leq y \leq \text{ln}\ x,0 \leq z \leq 1} \right\}$

193.

${\iiint\limits_{E}\left( {\text{sin}\ x + \text{sin}\ y} \right)}dV,$ where $E = \left\{ {\left. \left( {x,y,z} \right) \right|0 \leq x \leq \frac{\pi}{2},\text{−}\text{cos}\ x \leq y \leq \text{cos}\ x,-1 \leq z \leq 1} \right\}$

194\.

${\iiint\limits_{E}\left( {xy + yz + xz} \right)}dV,$ where $E = \left\{ {\left. \left( {x,y,z} \right) \right|0 \leq x \leq 1,\text{−}x^{2} \leq y \leq x^{2},0 \leq z \leq 1} \right\}$

In the following exercises, evaluate the triple integrals over the indicated bounded region $E.$

195.

${\iiint\limits_{E}{\left( {x + 2yz} \right)dV}},$ where $E = \left\{ (x,y,z) \middle| 0 \leq x \leq 1,0 \leq y \leq x,0 \leq z \leq 5 - x - y \right\}$

196\.

${\iiint\limits_{E}\left( {x^{3} + y^{3} + z^{3}} \right)}dV,$ where $E = \left\{ (x,y,z) \middle| 0 \leq x \leq 2,0 \leq y \leq 2x,0 \leq z \leq 4 - x - y \right\}$

197.

${\iiint\limits_{E}{y\ dV}},$ where $E = \left\{ (x,y,z) \middle| - 1 \leq x \leq 1,\text{−}\sqrt{1 - x^{2}} \leq y \leq \sqrt{1 - x^{2}},0 \leq z \leq 1 - x^{2} - y^{2} \right\}$

198\.

${\iiint\limits_{E}{x\ dV}},$ where $E = \left\{ (x,y,z) \middle| - 2 \leq x \leq 2, - \sqrt{4 - x^{2}} \leq y \leq \sqrt{4 - x^{2}},0 \leq z \leq 4 - x^{2} - y^{2} \right\}$

In the following exercises, evaluate the triple integrals over the bounded region $E$ of the form $E = \left\{ (x,y,z) \middle| g_{1}(y) \leq x \leq g_{2}(y),c \leq y \leq d,e \leq z \leq f \right\}.$

199.

${\iiint\limits_{E}{x^{2}dV}},$ where $E = \left\{ (x,y,z) \middle| 1 - y^{2} \leq x \leq y^{2} - 1,-1 \leq y \leq 1,1 \leq z \leq 2 \right\}$

200\.

${\iiint\limits_{E}\left( {\text{sin}\ x + y} \right)}dV,$ where $E = \left\{ (x,y,z) \middle| - y^{4} \leq x \leq y^{4},0 \leq y \leq 2,0 \leq z \leq 4 \right\}$

201.

${\iiint\limits_{E}\left( {x - yz} \right)}dV,$ where $E = \left\{ (x,y,z) \middle| - y^{6} \leq x \leq \sqrt{y},0 \leq y \leq 1,-1 \leq z \leq 1 \right\}$

202\.

${\iiint\limits_{E}z}dV,$ where $E = \left\{ (x,y,z) \middle| 2 - 2y \leq x \leq 2 + \sqrt{y},0 \leq y \leq 1,2 \leq z \leq 3 \right\}$

In the following exercises, evaluate the triple integrals over the bounded region

$$E = \left\{ (x,y,z) \middle| g_{1}(y) \leq x \leq g_{2}(y),c \leq y \leq d,u_{1}\left( {x,y} \right) \leq z \leq u_{2}\left( {x,y} \right) \right\}.$$ 203.

${\iiint\limits_{E}z}dV,$ where $E = \left\{ (x,y,z) \middle| - y \leq x \leq y,0 \leq y \leq 1,0 \leq z \leq 1 - x^{4} - y^{4} \right\}$

204\.

${\iiint\limits_{E}\left( {xz + 1} \right)}dV,$ where $E = \left\{ (x,y,z) \middle| 0 \leq x \leq \sqrt{y},0 \leq y \leq 2,0 \leq z \leq 1 - x^{2} - y^{2} \right\}$

205.

${\iiint\limits_{E}\left( {x - z} \right)}dV,$ where $E = \left\{ (x,y,z) \middle| - \sqrt{1 - y^{2}} \leq x \leq 0,0 \leq y \leq \frac{1}{2},0 \leq z \leq 1 - x^{2} - y^{2} \right\}$

206\.

${\iiint\limits_{E}\left( {x + y} \right)}dV,$ where $E = \left\{ (x,y,z) \middle| 0 \leq x \leq \sqrt{1 - y^{2}},0 \leq y \leq 1,0 \leq z \leq 1 - x \right\}$

In the following exercises, evaluate the triple integrals over the bounded region

$E = \left\{ (x,y,z) \middle| (x,y) \in D,u_{1}(x,y) \leq z \leq u_{2}(x,y) \right\},$ where $D$ is the projection of $E$ onto the $xy$-plane.

207.

${\iint\limits_{D}\left( {\int\limits_{1}^{2}{\left( {x + z} \right)dz}} \right)}dA,$ where $D = \left\{ (x,y) \middle| x^{2} + y^{2} \leq 1 \right\}$

208\.

${\iint\limits_{D}\left( {\int\limits_{1}^{3}{x\left( {z + 1} \right)dz}} \right)}dA,$ where $D = \left\{ (x,y) \middle| x^{2} - y^{2} \geq 1,1 \leq x \leq \sqrt{5} \right\}$

209.

${\iint\limits_{D}\left( {\int\limits_{0}^{10 - x - y}{\left( {x + 2z} \right)dz}} \right)}dA,$ where $D = \left\{ (x,y) \middle| y \geq 0,x \geq 0,x + y \leq 10 \right\}$

210\.

${\iint\limits_{D}\left( {\int\limits_{0}^{4x^{2} + 4y^{2}}{y\ dz}} \right)}dA,$ where $D = \left\{ (x,y) \middle| x^{2} + y^{2} \leq 4,y \geq 1,x \geq 0 \right\}$

211.

The solid $E$ bounded by $y^{2} + z^{2} = 9,z = 0,x = 0,$ and $x = 5$ is shown in the following figure. Evaluate the integral $\iiint\limits_{E}{z\ dV}$ by integrating first with respect to $z,$ then $y,\ \text{and then}\ x.$

212\.

The solid $E$ bounded by $y = \sqrt{x},$ $x = 4,$ $y = 0,$ $z = - 2$, and $z = 1$ is given in the following figure. Evaluate the integral $\iiint\limits_{E}{xyz\ dV}$ by integrating first with respect to $x,$ then $y,$ and then $z.$

213.

\[T\] The volume of a solid $E$ is given by the integral ${\int\limits_{-2}^{0}\ {\int\limits_{x}^{0}\ {\int\limits_{0}^{x^{2} + y^{2}}{dz\ dy\ dx}}}}.$ Use a computer algebra system (CAS) to graph $E$ and find its volume. Round your answer to two decimal places.

214\.

\[T\] The volume of a solid $E$ is given by the integral ${\int\limits_{-1}^{0}\ {\int\limits_{\text{−}x^{2}}^{0}\ {\int\limits_{0}^{1 + \sqrt{x^{2} + y^{2}}}{dz\ dy\ dx}}}}.$ Use a CAS to graph $E$ and find its volume $V.$ Round your answer to two decimal places.

In the following exercises, use two circular permutations of the variables $x,y,\ \text{and}\ z$ to write new integrals whose values equal the value of the original integral. A circular permutation of $x,y,\ \text{and}\ z$ is the arrangement of the numbers in one of the following orders: $y,z,\ \text{and}\ x\ \text{or}\ z,x,\ \text{and}\ y.$

215.

${\int\limits_{0}^{1}\ {\int\limits_{1}^{3}\ {\int\limits_{2}^{4}\left( {x^{2}z^{2} + 1} \right)}}}dx\ dy\ dz$

216\.

$\int\limits_{1}^{3}\ \int\limits_{0}^{1}\ \int\limits_{0}^{\text{−y} + 1}(2y + 5z + 7x)dz\ dy\ dx$

217.

${\int\limits_{0}^{1}\ {\int\limits_{\text{−}y}^{y}\ {\int\limits_{0}^{1 - x^{4} - y^{4}}e^{x}}}}\ dz\ dx\ dy$

218\.

${\int\limits_{-1}^{1}\ {\int\limits_{0}^{1}\ {\int\limits_{\text{−}y^{6}}^{\sqrt{y}}\left( {x + yz} \right)}}}dx\ dy\ dz$

219.

Set up the integral that gives the volume of the solid $E$ bounded by $y^{2} = x^{2} + z^{2}$ and $y = a,$ where $a > 0$ and $y~ \geq ~0$

220\.

Set up the integral that gives the volume of the solid $E$ bounded by $x = y^{2} + z^{2}$ and $x = a^{2},$ where $a > 0.$

221.

Find the average value of the function $f\left( {x,y,z} \right) = x + y + z$ over the parallelepiped determined by $x = 0,x = 1,y = 0,y = 3,z = 0,$ and $z = 5.$

222\.

Find the average value of the function $f\left( {x,y,z} \right) = xyz$ over the solid $E = \left\lbrack {0,1} \right\rbrack\ \times \ \left\lbrack {0,1} \right\rbrack\ \times \ \left\lbrack {0,1} \right\rbrack$ situated in the first octant.

223.

Find the volume of the solid $E$ that lies under the plane $x + y + z = 9$ and whose projection onto the $xy$-plane is bounded by $x = \sqrt{y - 1},x = 0,$ and $x + y = 7.$

224\.

Find the volume of the solid *E* that lies under the plane $2x + y + z = 8$ and whose projection onto the $xy$-plane is bounded by $x = \text{sin}^{-1}y,y = 0,$ and $x = \frac{\pi}{2}.$

225.

Consider the pyramid with the base in the $xy$-plane of $\left\lbrack {-2,2} \right\rbrack\ \times \ \left\lbrack {-2,2} \right\rbrack$ and the vertex at the point $\left( {0,0,8} \right).$

1. Show that the equations of the planes of the lateral faces of the pyramid are $4y + z = 8,$ $4y - z = -8,$ $4x + z = 8,$ and $-4x + z = 8.$

2. Find the volume of the pyramid.

226\.

Consider the pyramid with the base in the $xy$-plane of $\left\lbrack {-3,3} \right\rbrack\ \times \ \left\lbrack {-3,3} \right\rbrack$ and the vertex at the point $\left( {0,0,9} \right).$

1. Show that the equations of the planes of the side faces of the pyramid are $$3x + z = 9, - 3x + z = 9, - 3y + z = 9,\text{~and~}3y + z = 9$$

2. Find the volume of the pyramid.

227.

The solid $E$ bounded by the sphere of equation $x^{2} + y^{2} + z^{2} = r^{2}$ with $r > 0$ and located in the first octant is represented in the following figure.

1. Write the triple integral that gives the volume of $E$ by integrating first with respect to $z,$ then with $y,$ and then with $x.$

2. Rewrite the integral in part a. as an equivalent integral in five other orders.

228\.

The solid $E$ bounded by the equation $9x^{2} + 4y^{2} + z^{2} = 1$ and located in the first octant is represented in the following figure.

1. Write the triple integral that gives the volume of $E$ by integrating first with respect to $z,$ then with $y,$ and then with $x.$

2. Rewrite the integral in part a. as an equivalent integral in five other orders.

229.

Find the volume of the prism with vertices $\left( {0,0,0} \right),\left( {2,0,0} \right),\left( {2,3,0} \right),$ $\left( {0,3,0} \right),\left( {0,0,1} \right),\ \text{and}\ \left( {2,0,1} \right).$

230\.

Find the volume of the prism with vertices $\left( {0,0,0} \right),\left( {4,0,0} \right),\left( {4,6,0} \right),$ $\left( {0,6,0} \right),\left( {0,0,1} \right),\ \text{and}\ \left( {4,0,1} \right).$

231.

The solid $E$ bounded by $z = 10 - 2x - y$ and situated in the first octant is given in the following figure. Find the volume of the solid.

232\.

The solid $E$ bounded by $z = 1 - x^{2}$ and $y = 5$ and situated in the first octant is given in the following figure. Find the volume of the solid.

233.

The midpoint rule for the triple integral $\iiint\limits_{B}{f\left( {x,y,z} \right)dV}$ over the rectangular solid box $B$ is a generalization of the midpoint rule for double integrals. The region $B$ is divided into subboxes of equal sizes and the integral is approximated by the triple Riemann sum ${\sum\limits_{i = 1}^{l}{\sum\limits_{j = 1}^{m}{\sum\limits_{k = 1}^{n}{f\left( {\overset{–}{x_{i}},\overset{–}{y_{j}},\overset{–}{z_{k}}} \right)}}}}\text{Δ}V,$ where $\left( {\overset{–}{x_{i}},\overset{–}{y_{j}},\overset{–}{z_{k}}} \right)$ is the center of the box $B_{ijk}$ and $\text{Δ}V$ is the volume of each subbox. Apply the midpoint rule to approximate $\iiint\limits_{B}{x^{2}dV}$ over the solid $B = \left\{ {\left. \left( {x,y,z} \right) \right|0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1} \right\}$ by using a partition of eight cubes of equal size. Round your answer to three decimal places.

234\.

\[T\]

1. Apply the midpoint rule to approximate $\iiint\limits_{B}{e^{\text{−}x^{2}}dV}$ over the solid $B = \left\{ (x,y,z) \middle| 0 \leq x \leq 1,0 \leq y \leq 1,0 \leq z \leq 1 \right\}$ by using a partition of eight cubes of equal size. Round your answer to three decimal places.

2. Use a CAS to improve the above integral approximation in the case of a partition of $n^{3}$ cubes of equal size, where $n = 3,4\text{,…,}\ 10.$

235.

Suppose that the temperature in degrees Celsius at a point $\left( {x,y,z} \right)$ of a solid $E$ bounded by the coordinate planes and $x + y + z = 5$ is $T\left( {x,y,z} \right) = xz + 5z + 10.$ Find the average temperature over the solid.

236\.

Suppose that the temperature in degrees Fahrenheit at a point $\left( {x,y,z} \right)$ of a solid $E$ bounded by the coordinate planes and $x + y + z = 5$ is $T\left( {x,y,z} \right) = x + y + xy.$ Find the average temperature over the solid.

237\.

Show that the volume of a right square pyramid of height $h$ and side length $a$ is $v = \frac{ha^{2}}{3}$ by using triple integrals.

238\.

Show that the volume of a regular right hexagonal prism of edge length $a$ is $\frac{3a^{3}\sqrt{3}}{2}$ by using triple integrals.

239\.

Show that the volume of a regular right hexagonal pyramid of edge length $a$ is $\frac{a^{3}\sqrt{3}}{2}$ by using triple integrals.

240\.

If the charge density at an arbitrary point $(x,y,z)$ of a solid $E$ is given by the function $\rho(x,y,z),$ then the total charge inside the solid is defined as the triple integral ${\iiint\limits_{E}{\rho(x,y,z)dV}}.$ Assume that the charge density of the solid $E$ enclosed by the paraboloids $x = 5 - y^{2} - z^{2}$ and $x = y^{2} + z^{2} - 5$ is equal to the distance from an arbitrary point of $E$ to the origin. Set up the integral that gives the total charge inside the solid $E.$

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5.5 Triple Integrals in Cylindrical and Spherical Coordinates 5.5 柱坐标与球坐标下的三重积分

• 5.5.1 Evaluate a triple integral by changing to cylindrical coordinates.
• 5.5.2 Evaluate a triple integral by changing to spherical coordinates.

• 5.5.1 通过化为柱坐标来计算三重积分。
• 5.5.2 通过化为球坐标来计算三重积分。

Earlier in this chapter we showed how to convert a double integral in rectangular coordinates into a double integral in polar coordinates in order to deal more conveniently with problems involving circular symmetry. A similar situation occurs with triple integrals, but here we need to distinguish between cylindrical symmetry and spherical symmetry. In this section we convert triple integrals in rectangular coordinates into a triple integral in either cylindrical or spherical coordinates.

Also recall the chapter opener, which showed the opera house l’Hemisphèric in Valencia, Spain. It has four sections with one of the sections being a theater in a five-story-high sphere (ball) under an oval roof as long as a football field. Inside is an IMAX screen that changes the sphere into a planetarium with a sky full of $9000$ twinkling stars. Using triple integrals in spherical coordinates, we can find the volumes of different geometric shapes like these.

Review of Cylindrical Coordinates 柱坐标回顾

As we have seen earlier, in two-dimensional space $\mathbb{R}^{2},$ a point with rectangular coordinates $\left( {x,y} \right)$ can be identified with $\left( {r,\theta} \right)$ in polar coordinates and vice versa, where $x = r\ \text{cos}\ \theta,$ $y = r\ \text{sin}\ \theta,$ $r^{2} = x^{2} + y^{2}$ and $\text{tan}\ \theta = \left( \frac{y}{x} \right)$ are the relationships between the variables.

In three-dimensional space $\mathbb{R}^{3},$ a point with rectangular coordinates $\left( {x,y,z} \right)$ can be identified with cylindrical coordinates $\left( {r,\theta,z} \right)$ and vice versa. We can use these same conversion relationships, adding $z$ as the vertical distance to the point from the $xy$-plane as shown in the following figure.

To convert from rectangular to cylindrical coordinates, we use the conversion $x = r\ \text{cos}\ \theta$ and $y = r\ \text{sin}\ \theta.$ To convert from cylindrical to rectangular coordinates, we use $r^{2} = x^{2} + y^{2}$ and $\text{tan}\theta = \frac{y}{x}.$ The $z$-coordinate remains the same in both cases.

In the two-dimensional plane with a rectangular coordinate system, when we say $x = k$ (constant) we mean an unbounded vertical line parallel to the $y$-axis and when $y = l$ (constant) we mean an unbounded horizontal line parallel to the $x$-axis. With the polar coordinate system, when we say $r = c$ (constant), we mean a circle of radius $c$ units and when $\theta = \alpha$ (constant) we mean an infinite ray making an angle $\alpha$ with the positive $x$-axis.

Similarly, in three-dimensional space with rectangular coordinates $\left( {x,y,z} \right),$ the equations $x = k,y = l,$ and $z = m,$ where $k,l,$ and $m$ are constants, represent unbounded planes parallel to the $yz$-plane, $xz$-plane and $xy$-plane, respectively. With cylindrical coordinates $\left( {r,\theta,z} \right),$ by $r = c,\theta = \alpha,$ and $z = m,$ where $c,\alpha,$ and $m$ are constants, we mean an unbounded vertical cylinder with the $z$-axis as its radial axis; a plane making a constant angle $\alpha$ with the $xz$-plane; and an unbounded horizontal plane parallel to the $xy$-plane, respectively. This means that the circular cylinder $x^{2} + y^{2} = c^{2}$ in rectangular coordinates can be represented simply as $r = c$ in cylindrical coordinates. (Refer to Cylindrical and Spherical Coordinates for more review.)

Integration in Cylindrical Coordinates 柱坐标下的积分

Triple integrals can often be more readily evaluated by using cylindrical coordinates instead of rectangular coordinates. Some common equations of surfaces in rectangular coordinates along with corresponding equations in cylindrical coordinates are listed in Table 5.1. These equations will become handy as we proceed with solving problems using triple integrals.

| | Circular cylinder | Circular cone | Sphere | Paraboloid |

|-------------|-------------------------|-----------------------------------------------|---------------------------------|---------------------------------------|

| Rectangular | $x^{2} + y^{2} = c^{2}$ | $z^{2} = c^{2}\left( {x^{2} + y^{2}} \right)$ | $x^{2} + y^{2} + z^{2} = c^{2}$ | $z = c\left( {x^{2} + y^{2}} \right)$ |

| Cylindrical | $r = c$ | $z = cr$ | $r^{2} + z^{2} = c^{2}$ | $z = cr^{2}$ |

Table 5.1 Equations of Some Common Shapes

As before, we start with the simplest bounded region $B$ in $\mathbb{R}^{3},$ to describe in cylindrical coordinates, in the form of a cylindrical box, $B = \left\{ {\left. \left( {r,\theta,z} \right) \right|a \leq r \leq b,\alpha \leq \theta \leq \beta,c \leq z \leq d} \right\}$ (Figure 5.51). Suppose we divide each interval into $l,m\ \text{and}\ n$ subdivisions such that $\text{Δ}r = \frac{b - a}{l},\text{Δ}\theta = \frac{\beta - \alpha}{m},$ and $\text{Δ}z = \frac{d - c}{n}.$ Then we can state the following definition for a triple integral in cylindrical coordinates.

Consider the cylindrical box (expressed in cylindrical coordinates)

$$B = \left\{ {\left. \left( {r,\theta,z} \right) \right|a \leq r \leq b,\alpha \leq \theta \leq \beta,c \leq z \leq d} \right\}.$$

If the function $f\left( {r,\theta,z} \right)$ is continuous on $B$ and if $(r_{ijk}^{*},\theta_{ijk}^{*},z_{ijk}^{*})$ is any sample point in the cylindrical subbox $B_{ijk} = \left\lbrack {r_{i - 1},r_{i}} \right\rbrack\ \times \ \left\lbrack {\theta_{j - 1},\theta_{j}} \right\rbrack\ \times \ \left\lbrack {z_{k - 1},z_{k}} \right\rbrack$ (Figure 5.51), then we can define the triple integral in cylindrical coordinates as the limit of a triple Riemann sum, provided the following limit exists:

$$\underset{l,m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{l}{\sum\limits_{j = 1}^{m}{\sum\limits_{k = 1}^{n}{f(r_{ijk}^{*},\theta_{ijk}^{*},z_{ijk}^{*})r_{ijk}^{*}\text{Δ}r\text{Δ}\theta\text{Δ}z}}}}.$$

Note that if $g\left( {x,y,z} \right)$ is the function in rectangular coordinates and the box $B$ is expressed in rectangular coordinates, then the triple integral $\iiint\limits_{B}{g\left( {x,y,z} \right)dV}$ is equal to the triple integral $\iiint\limits_{B}{g\left( {r\ \text{cos}\ \theta,r\ \text{sin}\ \theta,z} \right)r\ dr\ d\theta\ dz}$ and we have

$${\iiint\limits_{B}{g\left( {x,y,z} \right)dV}} = {\iiint\limits_{B}{g\left( {r\ \text{cos}\ \theta,r\ \text{sin}\ \theta,z} \right)r\ dr\ d\theta\ dz}} = {\iiint\limits_{B}{f\left( {r,\theta,z} \right)r\ dr\ d\theta\ dz}}.$$ (5.11)

As mentioned in the preceding section, all the properties of a double integral work well in triple integrals, whether in rectangular coordinates or cylindrical coordinates. They also hold for iterated integrals. To reiterate, in cylindrical coordinates, Fubini’s theorem takes the following form:

Fubini’s Theorem in Cylindrical Coordinates 柱坐标下的富比尼定理

Suppose that $g\left( {x,y,z} \right)$ is continuous on a portion of a circular cylinder $B,$ which when described in cylindrical coordinates looks like $B = \left\{ {\left. \left( {r,\theta,z} \right) \right|a \leq r \leq b,\alpha \leq \theta \leq \beta,c \leq z \leq d} \right\}.$

Then $g\left( {x,y,z} \right) = g\left( {r\ \text{cos}\ \theta,r\ \text{sin}\ \theta,z} \right) = f\left( {r,\theta,z} \right)$ and

$${\iiint\limits_{B}{g\left( {x,y,z} \right)dV = {\int\limits_{c}^{d}\ {\int\limits_{\alpha}^{\beta}\ {\int\limits_{a}^{b}{f\left( {r,\theta,z} \right)}}}}}}r\ dr\ d\theta\ dz.$$

The iterated integral may be replaced equivalently by any one of the other five iterated integrals obtained by integrating with respect to the three variables in other orders.

Cylindrical coordinate systems work well for solids that are symmetric around an axis, such as cylinders and cones. Let us look at some examples before we define the triple integral in cylindrical coordinates on general cylindrical regions.

Evaluating a Triple Integral over a Cylindrical Box 在柱形盒上计算三重积分

Evaluate the triple integral ${\iiint\limits_{B}\left( {zr\ \text{sin}\ \theta} \right)}r\ dr\ d\theta\ dz$ where the cylindrical box $B$ is $B = \left\{ {\left. \left( {r,\theta,z} \right) \right|0 \leq r \leq 2,0 \leq \theta \leq \pi\text{/}2,0 \leq z \leq 4} \right\}.$

Solution 解

As stated in Fubini’s theorem, we can write the triple integral as the iterated integral

$${\iiint\limits_{B}\left( {zr\ \text{sin}\ \theta} \right)}r\ dr\ d\theta\ dz = {\int_{\theta = 0}^{\theta = \pi\text{/}2}{\int_{r = 0}^{r = 2}{\int_{z = 0}^{z = 4}\left( {zr\ \text{sin}\ \theta} \right)}}}r\ dz\ dr\ d\theta.$$

The evaluation of the iterated integral is straightforward. Each variable in the integral is independent of the others, so we can integrate each variable separately and multiply the results together. This makes the computation much easier:

$$\begin{array}{l} \\ \\ \\ \\ {\quad{\int_{\theta = 0}^{\theta = \pi\text{/}2}{\int_{r = 0}^{r = 2}{\int_{z = 0}^{z = 4}\left( {zr\ \text{sin}\ \theta} \right)}}}r\ dz\ dr\ d\theta} \\ {= \left( {\int_{0}^{\pi\text{/}2}{\text{sin}\ \theta\ d\theta}} \right)\left( {\int_{0}^{2}{r^{2}dr}} \right)\left( {\int_{0}^{4}{z\ dz}} \right) = \left( \left. {\text{−}\text{cos}\ \theta} \right|_{0}^{\pi\text{/}2} \right)\left( \left. \frac{r^{3}}{3} \right|_{0}^{2} \right)\left( \left. \frac{z^{2}}{2} \right|_{0}^{4} \right) = \frac{64}{3}.} \end{array}$$

Evaluate the triple integral $\int\limits_{\theta = 0}^{\theta = \pi}\ \int\limits_{r = 0}^{r = 1}\ \int\limits_{z = 0}^{z = 4}\left( rz\ \text{sin}\ \theta \right)rdzdrd\theta.$

If the cylindrical region over which we have to integrate is a general solid, we look at the projections onto the coordinate planes. Hence the triple integral of a continuous function $f\left( {r,\theta,z} \right)$ over a general solid region $E = \left\{ {\left. \left( {r,\theta,z} \right) \right|\left( {r,\theta} \right) \in D,u_{1}\left( {r,\theta} \right) \leq z \leq u_{2}\left( {r,\theta} \right)} \right\}$ in $\mathbb{R}^{3},$ where $D$ is the projection of $E$ onto the $r\theta$-plane, is

$${\iiint\limits_{E}{f\left( {r,\theta,z} \right)r\ dr\ d\theta\ dz = {\iint\limits_{D}\left\lbrack {\int\limits_{u_{1}{({r,\theta})}}^{u_{2}{({r,\theta})}}{f\left( {r,\theta,z} \right)dz}} \right\rbrack}}}r\ dr\ d\theta.$$

In particular, if $D = \left\{ {\left. \left( {r,\theta} \right) \right|g_{1}(\theta) \leq r \leq g_{2}(\theta),\alpha \leq \theta \leq \beta} \right\},$ then we have

$${\iiint\limits_{E}{f\left( {r,\theta,z} \right)}}r\ dr\ d\theta = {\int\limits_{\theta = \alpha}^{\theta = \beta}\ {\int\limits_{r = g_{1}{(\theta)}}^{r = g_{2}{(\theta)}}\ {\int\limits_{z = u_{1}{({r,\theta})}}^{z = u_{2}{({r,\theta})}}{f\left( {r,\theta,z} \right)}}}}r\ dz\ dr\ d\theta.$$

Similar formulas exist for projections onto the other coordinate planes. We can use polar coordinates in those planes if necessary.

Setting up a Triple Integral in Cylindrical Coordinates over a General Region 在一般区域上建立柱坐标三重积分

Consider the region $E$ inside the right circular cylinder with equation $r = 2\ \text{sin}\ \theta,$ bounded below by the $r\theta$-plane and bounded above by the sphere with radius $4$ centered at the origin (Figure 5.52). Set up a triple integral over this region with a function $f\left( {r,\theta,z} \right)$ in cylindrical coordinates.

Solution 解

First, identify that the equation for the sphere is $r^{2} + z^{2} = 16.$ We can see that the limits for $z$ are from $0$ to $z = \sqrt{16 - r^{2}}.$ Then the limits for $r$ are from $0$ to $r = 2\ \text{sin}\ \theta.$ Finally, the limits for $\theta$ are from $0$ to $\pi.$ Hence the region is

$$E = \left\{ {\left. \left( {r,\theta,z} \right) \right|0 \leq \theta \leq \pi,0 \leq r \leq 2\ \text{sin}\ \theta,0 \leq z \leq \sqrt{16 - r^{2}}} \right\}.$$

Therefore, the triple integral is

$${\iiint\limits_{E}{f\left( {r,\theta,z} \right)}}r\ dz\ dr\ d\theta = {\int\limits_{\theta = 0}^{\theta = \pi}\ {\int\limits_{r = 0}^{r = 2\ \text{sin}\ \theta}\ {\int\limits_{z = 0}^{z = \sqrt{16 - r^{2}}}{f\left( {r,\theta,z} \right)}}}}r\ dz\ dr\ d\theta.$$

Consider the region $E$ inside the right circular cylinder with equation $r = 2\ \text{sin}\ \theta,$ bounded below by the $r\theta$-plane and bounded above by $z = 4 - y.$ Set up a triple integral with a function $f\left( {r,\theta,z} \right)$ in cylindrical coordinates.

Setting up a Triple Integral in Two Ways 用两种方式建立三重积分

Let $E$ be the region bounded below by the cone $z = \sqrt{x^{2} + y^{2}}$ and above by the paraboloid $z = 2 - x^{2} - y^{2}.$ (Figure 5.53). Set up a triple integral in cylindrical coordinates to find the volume of the region, using the following orders of integration:

1. $dz\ dr\ d\theta$

2. $dr\ dz\ d\theta.$

Solution 解

1. The cone is of radius 1 where it meets the paraboloid. Since $z = 2 - x^{2} - y^{2} = 2 - r^{2}$ and $z = \sqrt{x^{2} + y^{2}} = r$ (assuming $r$ is nonnegative), we have $2 - r^{2} = r.$ Solving, we have $r^{2} + r - 2 = \left( {r + 2} \right)\left( {r - 1} \right) = 0.$ Since $r \geq 0,$ we have $r = 1.$ Therefore $z = 1.$ So the intersection of these two surfaces is a circle of radius $1$ in the plane $z = 1.$ The cone is the lower bound for $z$ and the paraboloid is the upper bound. The projection of the region onto the $xy$-plane is the circle of radius $1$ centered at the origin.

Thus, we can describe the region as

$$E = \left\{ {\left. \left( {r,\theta,z} \right) \right|0 \leq \theta \leq 2\pi,0 \leq r \leq 1,r \leq z \leq 2 - r^{2}} \right\}.$$

Hence the integral for the volume is

$$V = {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{r = 0}^{r = 1}\ {\int\limits_{z = r}^{z = 2 - r^{2}}{r\ dz\ dr\ d\theta}}}}.$$

2. We can also write the cone surface as $r = z$ and the paraboloid as $r^{2} = 2 - z.$ The lower bound for $r$ is zero, but the upper bound is sometimes the cone and the other times it is the paraboloid. The plane $z = 1$ divides the region into two regions. Then the region can be described as

$$\begin{array}{cl} E & {= \left\{ {\left. \left( {r,\theta,z} \right) \right|0 \leq \theta \leq 2\pi,0 \leq z \leq 1,0 \leq r \leq z} \right\}} \\ & {\cup \left\{ {\left. \left( {r,\theta,z} \right) \right|0 \leq \theta \leq 2\pi,1 \leq z \leq 2,0 \leq r \leq \sqrt{2 - z}} \right\}.} \end{array}$$

Now the integral for the volume becomes

$$V = {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{z = 0}^{z = 1}\ {\int\limits_{r = 0}^{r = z}{r\ dr\ dz\ d\theta +}}}}{\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{z = 1}^{z = 2}\ {\int\limits_{r = 0}^{r = \sqrt{2 - z}}{r\ dr\ dz\ d\theta}}}}.$$

Redo the previous example with the order of integration $d\theta\ dz\ dr.$

Finding a Volume with Triple Integrals in Two Ways 用两种方法求三重积分的体积

Let *E* be the region bounded below by the $r\theta$-plane, above by the sphere $x^{2} + y^{2} + z^{2} = 4,$ and on the sides by the cylinder $x^{2} + y^{2} = 1$ (Figure 5.54). Set up a triple integral in cylindrical coordinates to find the volume of the region using the following orders of integration, and in each case find the volume and check that the answers are the same:

1. $dz\ dr\ d\theta$

2. $dr\ dz\ d\theta.$

Solution 解

1. Note that the equation for the sphere is

$$x^{2} + y^{2} + z^{2} = 4\ \text{or}\ r^{2} + z^{2} = 4$$

and the equation for the cylinder is

$$x^{2} + y^{2} = 1\ \text{or}\ r^{2} = 1.$$

Thus, we have for the region $E$

$$E = \left\{ {\left. \left( {r,\theta,z} \right) \right|0 \leq z \leq \sqrt{4 - r^{2}},0 \leq r \leq 1,0 \leq \theta \leq 2\pi} \right\}$$

Hence the integral for the volume is

$$\begin{array}{cl} {V(E)} & {= {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{r = 0}^{r = 1}\ {\int\limits_{z = 0}^{z = \sqrt{4 - r^{2}}}{r\ dz\ dr\ d\theta}}}}} \\ & {= {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{r = 0}^{r = 1}\left\lbrack \left. {rz} \right|_{z = 0}^{z = \sqrt{4 - r^{2}}} \right\rbrack}}dr\ d\theta = {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{r = 0}^{r = 1}\left( {r\sqrt{4 - r^{2}}} \right)}}dr\ d\theta} \\ & {= {\int\limits_{0}^{2\pi}{\left( {\frac{8}{3} - \sqrt{3}} \right)d\theta = 2\pi\left( {\frac{8}{3} - \sqrt{3}} \right)}}\ \text{cubic units}\text{.}} \end{array}$$

2. Since the sphere is $x^{2} + y^{2} + z^{2} = 4,$ which is $r^{2} + z^{2} = 4,$ and the cylinder is $x^{2} + y^{2} = 1,$ which is $r^{2} = 1,$ we have $1 + z^{2} = 4,$ that is, $z^{2} = 3.$ Thus we have two regions, since the sphere and the cylinder intersect at $\left( {1,\sqrt{3}} \right)$ in the $rz$-plane

$$E_{1} = \left\{ {\left. \left( {r,\theta,z} \right) \right|0 \leq r \leq \sqrt{4 - z^{2}},\sqrt{3} \leq z \leq 2,0 \leq \theta \leq 2\pi} \right\}$$

and

$$E_{2} = \left\{ {\left. \left( {r,\theta,z} \right) \right|0 \leq r \leq 1,0 \leq z \leq \sqrt{3},0 \leq \theta \leq 2\pi} \right\}.$$

Hence the integral for the volume is

$$\begin{array}{cl} {V(E)} & {= {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{z = \sqrt{3}}^{z = 2}\ {\int\limits_{r = 0}^{r = \sqrt{4 - r^{2}}}{r\ dr\ dz\ d\theta}}}} + {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{z = 0}^{z = \sqrt{3}}\ {\int\limits_{r = 0}^{r = 1}{r\ dr\ dz\ d\theta}}}}} \\ & {= \sqrt{3}\pi + \left( {\frac{16}{3} - 3\sqrt{3}} \right)\pi = 2\pi\left( {\frac{8}{3} - \sqrt{3}} \right)\ \text{cubic units}.} \end{array}$$

Redo the previous example with the order of integration $d\theta\ dz\ dr.$

Review of Spherical Coordinates 球坐标回顾

In three-dimensional space $\mathbb{R}^{3}$ in the spherical coordinate system, we specify a point $P$ by its distance $\rho$ from the origin, the polar angle $\theta$ from the positive $x\text{-axis}$ (same as in the cylindrical coordinate system), and the angle $\varphi$ from the positive $z\text{-axis}$ and the line $OP$ (Figure 5.55). Note that $\rho \geq 0$ and $0 \leq \varphi \leq \pi.$ (Refer to Cylindrical and Spherical Coordinates for a review.) Spherical coordinates are useful for triple integrals over regions that are symmetric with respect to the origin.

Recall the relationships that connect rectangular coordinates with spherical coordinates.

From spherical coordinates to rectangular coordinates:

$$x = \rho\ \text{sin}\ \varphi\ \text{cos}\ \theta,y = \rho\ \text{sin}\ \varphi\ \text{sin}\ \theta,\ \text{and}\ z = \rho\ \text{cos}\ \varphi.$$

From rectangular coordinates to spherical coordinates:

$$\rho^{2} = x^{2} + y^{2} + z^{2},\text{tan}\ \theta = \frac{y}{x},\varphi = \text{arccos}\left( \frac{z}{\sqrt{x^{2} + y^{2} + z^{2}}} \right).$$

Other relationships that are important to know for conversions are

$$\begin{array}{lclccc} \text{•} & & {r = \rho\ \text{sin}\ \varphi} & & & \\ \text{•} & & {\theta = \theta} & & & \begin{array}{l} \text{These equations are used to convert from} \\ \text{spherical coordinates to cylindrical coordinates} \end{array} \\ \text{•} & & {z = \rho\ \text{cos}\ \varphi} & & & \end{array}$$

and

$$\begin{array}{lclccc} & & & & & \\ & & & & & \\ \text{•} & & {\rho = \sqrt{r^{2} + z^{2}}} & & & \\ \text{•} & & {\theta = \theta} & & & \begin{array}{l} \text{These equations are used to convert from} \\ \text{cylindrical coordinates to spherical} \\ \text{coordinates.} \end{array} \\ \text{•} & & {\varphi = \text{arccos}\left( \frac{z}{\sqrt{r^{2} + z^{2}}} \right)} & & & \end{array}$$

The following figure shows a few solid regions that are convenient to express in spherical coordinates.

Integration in Spherical Coordinates 球坐标下的积分

We now establish a triple integral in the spherical coordinate system, as we did before in the cylindrical coordinate system. Let the function $f\left( {\rho,\theta,\varphi} \right)$ be continuous in a bounded spherical box, $B = \left\{ {\left. \left( {\rho,\theta,\varphi} \right) \right|a \leq \rho \leq b,\alpha \leq \theta \leq \beta,\gamma \leq \varphi \leq \psi} \right\}.$ We then divide each interval into $l,m\ \text{and}\ n$ subdivisions such that $\text{Δ}\rho = \frac{b - a}{l},\text{Δ}\theta = \frac{\beta - \alpha}{m},\text{Δ}\varphi = \frac{\psi - \gamma}{n}.$

Now we can illustrate the following theorem for triple integrals in spherical coordinates with $(\rho_{ijk}^{*},\theta_{ijk}^{*},\varphi_{ijk}^{*})$ being any sample point in the spherical subbox $B_{ijk}.$ For the volume element of the subbox $\text{Δ}V$ in spherical coordinates, we have $\text{Δ}V = \left( {\text{Δ}\rho} \right)\left( {\rho\text{Δ}\varphi} \right)\left( {\rho\ \text{sin}\ \varphi\text{Δ}\theta} \right),,$ as shown in the following figure.

The triple integral in spherical coordinates is the limit of a triple Riemann sum,

$$\underset{l,m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{l}{\sum\limits_{j = 1}^{m}{\sum\limits_{k = 1}^{n}{f(\rho_{ijk}^{*},\theta_{ijk}^{*},\varphi_{ijk}^{*}){(\rho_{ijk}^{*})}^{2}\text{sin}\ \varphi_{ijk}^{*}\text{Δ}\rho\text{Δ}\theta\text{Δ}\varphi}}}}$$

provided the limit exists.

As with the other multiple integrals we have examined, all the properties work similarly for a triple integral in the spherical coordinate system, and so do the iterated integrals. Fubini’s theorem takes the following form.

Fubini’s Theorem for Spherical Coordinates 球坐标下的富比尼定理

If $f\left( {\rho,\theta,\varphi} \right)$ is continuous on a spherical solid box $B = \left\lbrack {a,b} \right\rbrack\ \times \ \left\lbrack {\alpha,\beta} \right\rbrack\ \times \ \left\lbrack {\gamma,\psi} \right\rbrack,$ then

$$\iiint\limits_{B}f(\rho,\theta,\varphi)\rho^{2}\text{sin}\ \varphi\ d\rho\ d\varphi\ d\theta = \int\limits_{\theta = \alpha}^{\theta = \beta}~\int\limits_{\varphi = \gamma}^{\varphi = \psi}\ \int\limits_{\rho = a}^{\rho = b}f(\rho,\theta,\varphi)\rho^{2}\text{sin}\ \varphi\ d\rho\ d\varphi\ d\theta.$$ (5.12)

This iterated integral may be replaced by other iterated integrals by integrating with respect to the three variables in other orders.

As stated before, spherical coordinate systems work well for solids that are symmetric around a point, such as spheres and cones. Let us look at some examples before we consider triple integrals in spherical coordinates on general spherical regions.

Evaluating a Triple Integral in Spherical Coordinates 计算球坐标下的三重积分

Evaluate the iterated triple integral ${\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{\varphi = 0}^{\varphi = \pi\text{/}2}\ {\int\limits_{p = 0}^{\rho = 1}{\rho^{2}\text{sin}\ \varphi\ d\rho\ d\varphi\ d\theta}}}}.$

Solution 解

As before, in this case the variables in the iterated integral are actually independent of each other and hence we can integrate each piece and multiply:

$${\int\limits_{0}^{2\pi}\ {\int\limits_{0}^{\pi\text{/}2}\ {\int\limits_{0}^{1}{\rho^{2}\text{sin}\ \varphi\ d\rho\ d\varphi\ d\theta}}}} = {\int\limits_{0}^{2\pi}{d\theta}}{\int\limits_{0}^{\pi\text{/}2}{\text{sin}\ \varphi\ d\varphi{\int\limits_{0}^{1}{\rho^{2}d\rho}}}} = \left( {2\pi} \right)(1)\left( \frac{1}{3} \right) = \frac{2\pi}{3}.$$

The concept of triple integration in spherical coordinates can be extended to integration over a general solid, using the projections onto the coordinate planes. Note that $dV$ and $dA$ mean the increments in volume and area, respectively. The variables $V$ and $A$ are used as the variables for integration to express the integrals.

The triple integral of a continuous function $f\left( {\rho,\theta,\varphi} \right)$ over a general solid region

$$E = \left\{ {\left. \left( {\rho,\theta,\varphi} \right) \right|\left( {\rho,\theta} \right) \in D,u_{1}\left( {\rho,\theta} \right) \leq \varphi \leq u_{2}\left( {\rho,\theta} \right)} \right\}$$

in $\mathbb{R}^{3},$ where $D$ is the projection of $E$ onto the $\rho\theta$-plane, is

$${\iiint\limits_{E}{f\left( {\rho,\theta,\varphi} \right)}}dV = {\iint\limits_{D}\left\lbrack {\int\limits_{u_{1}{({\rho,\theta})}}^{u_{2}{({\rho,\theta})}}{f\left( {\rho,\theta,\varphi} \right)d\varphi}} \right\rbrack}dA.$$

In particular, if $D = \left\{ {\left. \left( {\rho,\theta} \right) \right|g_{1}(\theta) \leq \rho \leq g_{2}(\theta),\alpha \leq \theta \leq \beta} \right\},$ then we have

$${\iiint\limits_{E}{f\left( {\rho,\theta,\varphi} \right)}}dV = {\int\limits_{\alpha}^{\beta}\ {\int\limits_{g_{1}{(\theta)}}^{g_{2}{(\theta)}}\ {\int\limits_{u_{1}{({\rho,\theta})}}^{u_{2}{({\rho,\theta})}}{f\left( {\rho,\theta,\varphi} \right)\rho^{2}\text{sin}\ \varphi\ d\varphi\ d\rho\ d\theta}}}}.$$

Similar formulas occur for projections onto the other coordinate planes.

Setting up a Triple Integral in Spherical Coordinates 建立球坐标下的三重积分

Set up an integral for the volume of the region bounded by the cone $z = \sqrt{3\left( {x^{2} + y^{2}} \right)}$ and the hemisphere $z = \sqrt{4 - x^{2} - y^{2}}$ (see the figure below).

Solution 解

Using the conversion formulas from rectangular coordinates to spherical coordinates, we have:

For the cone: $z = \sqrt{3\left( {x^{2} + y^{2}} \right)}$ or $\rho\ \text{cos}\ \varphi = \sqrt{3}\rho\ \text{sin}\ \varphi$ or $\text{tan}\ \varphi = \frac{1}{\sqrt{3}}$ or $\varphi = \frac{\pi}{6}.$

For the sphere: $z = \sqrt{4 - x^{2} - y^{2}}$ or $z^{2} = x^{2} + y^{2} = 4$ or $\rho^{2} = 4$ or $\rho = 2.$

Thus, the triple integral for the volume is $V(E) = {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{\varphi = 0}^{\varphi = \pi\text{/}6}\ {\int\limits_{\rho = 0}^{\rho = 2}{\rho^{2}\text{sin}\ \varphi\ d\rho\ d\varphi\ d\theta}}}}.$

Set up a triple integral for the volume of the solid region bounded above by the sphere $\rho = 2$ and bounded below by the cone $\varphi = \pi\text{/}3.$

Interchanging Order of Integration in Spherical Coordinates 交换球坐标下的积分次序

Let $E$ be the region bounded below by the cone $z = \sqrt{x^{2} + y^{2}}$ and above by the sphere $z = x^{2} + y^{2} + z^{2}$ (Figure 5.59). Set up a triple integral in spherical coordinates and find the volume of the region using the following orders of integration:

1. $d\rho\ d\theta\ d\theta,$

2. $d\varphi\ d\rho\ d\theta.$

Solution 解

1. Use the conversion formulas to write the equations of the sphere and cone in spherical coordinates.

For the sphere:

$$\begin{array}{rll} {x^{2} + y^{2} + z^{2}} & = & z \\ \rho^{2} & = & {\rho\ \text{cos}\ \varphi} \\ \rho & = & {\text{cos}\ \varphi.} \end{array}$$

For the cone:

$$\begin{matrix} z & = & \sqrt{x^{2} + y^{2}} \\ {\rho\ \text{cos}\ \varphi} & = & \sqrt{\rho^{2}\text{sin}^{2}\varphi\ \text{cos}^{2}\text{θ} + \rho^{2}\text{sin}^{2}\varphi\ \text{sin}^{2}\text{θ}} \\ {\rho\ \text{cos}\ \varphi} & = & \sqrt{\rho^{2}\text{sin}^{2}\varphi\left( \text{cos}^{2}\text{θ} + \text{sin}^{2}\text{θ} \right)} \\ {\rho\ \text{cos}\ \varphi} & = & {\rho\ \text{sin}\ \varphi} \\ {\text{cos}\ \varphi} & = & {\text{sin}\ \varphi} \\ \varphi & = & {\pi\text{/}4.} \end{matrix}$$

Hence the integral for the volume of the solid region $E$ becomes

$$V(E) = {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{\varphi = 0}^{\varphi = \pi\text{/}4}\ {\int\limits_{\rho = 0}^{\rho = \text{cos}\ \varphi}{\rho^{2}\text{sin}\ \varphi\ d\rho\ d\varphi\ d\theta}}}}.$$

2. Consider the $\varphi\rho$-plane. Note that the ranges for $\varphi$ and $\rho$ (from part a.) are

$$\begin{array}{l} {0 \leq \varphi \leq \pi\text{/}4} \\ {0 \leq \rho \leq \text{cos}\ \varphi.} \end{array}$$

The curve $\rho = \text{cos}\ \varphi$ meets the line $\varphi = \pi\text{/}4$ at the point $\left( {\pi\text{/}4,\sqrt{2}\text{/}2} \right).$ Thus, to change the order of integration, we need to use two pieces:

$$\begin{array}{lcclccl} \begin{array}{l} {0 \leq \rho \leq \sqrt{2}\text{/}2} \\ {0 \leq \varphi \leq \pi\text{/}4} \end{array} & & & \text{and} & & & \begin{array}{rll} {\sqrt{2}\text{/}2} & \leq & {\rho \leq 1} \\ 0 & \leq & {\varphi \leq \text{cos}^{-1}\rho.} \end{array} \end{array}$$

Hence the integral for the volume of the solid region $E$ becomes

$$V(E) = {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{\rho = 0}^{\rho = \sqrt{2}\text{/}2}\ {\int\limits_{\varphi = 0}^{\varphi = \pi\text{/}4}{\rho^{2}\text{sin}\ \varphi\ d\varphi\ d\rho\ d\theta}}}} + {\int\limits_{\theta = 0}^{\theta = 2\pi}\ {\int\limits_{\rho = \sqrt{2}\text{/}2}^{\rho = 1}\ {\int\limits_{\varphi = 0}^{\varphi = \text{cos}^{-1}\rho}{\rho^{2}\text{sin}\ \varphi\ d\varphi\ d\rho\ d\theta}}}}.$$

In each case, the integration results in $V(E) = \frac{\pi}{8}.$

Before we end this section, we present a couple of examples that can illustrate the conversion from rectangular coordinates to cylindrical coordinates and from rectangular coordinates to spherical coordinates.

Converting from Rectangular Coordinates to Cylindrical Coordinates 从直角坐标化为柱坐标

Convert the following integral into cylindrical coordinates:

$${\int\limits_{y = -1}^{y = 1}\ {\int\limits_{x = 0}^{x = \sqrt{1 - y^{2}}}\ {\int\limits_{z = x^{2} + y^{2}}^{z = \sqrt{x^{2} + y^{2}}}{xyz\ dz\ dx\ dy}}}}.$$

Solution 解

The ranges of the variables are

$$\begin{array}{rll} {- 1} & \leq & {y \leq 1} \\ 0 & \leq & {x \leq \sqrt{1 - y^{2}}} \\ {x^{2} + y^{2}} & \leq & {z \leq \sqrt{x^{2} + y^{2}}.} \end{array}$$

The first two inequalities describe the right half of a circle of radius $1.$ Therefore, the ranges for $\theta$ and $r$ are

$$- \frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\ \text{and}\ 0 \leq r \leq 1.$$

The limits of $z$ are $r^{2} \leq z \leq r,$ hence

$${\int\limits_{y = -1}^{y = 1}\ {\int\limits_{x = 0}^{x = \sqrt{1 - y^{2}}}\ {\int\limits_{z = x^{2} + y^{2}}^{z = \sqrt{x^{2} + y^{2}}}{xyz\ dz\ dx\ dy}}}} = {\int\limits_{\theta = \text{−}{\pi\text{/}2}}^{\theta = {\pi\text{/}2}}\ {\int\limits_{r = 0}^{r = 1}\ {\int\limits_{z = r^{2}}^{z = r}{r\left( {r\ \text{cos}\ \theta} \right)\left( {r\ \text{sin}\ \theta} \right)z\ dz\ dr\ d\theta}}}}.$$

Converting from Rectangular Coordinates to Spherical Coordinates 从直角坐标化为球坐标

Convert the following integral into spherical coordinates:

$${\int\limits_{y = 0}^{y = 3}\ {\int\limits_{x = 0}^{x = \sqrt{9 - y^{2}}}\ {\int\limits_{z = \sqrt{x^{2} + y^{2}}}^{z = \sqrt{18 - x^{2} - y^{2}}}{\left( {x^{2} + y^{2} + z^{2}} \right)dz\ dx\ dy}}}}.$$

Solution 解

The ranges of the variables are

$$\begin{array}{rll} 0 & \leq & {y \leq 3} \\ 0 & \leq & {x \leq \sqrt{9 - y^{2}}} \\ \sqrt{x^{2} + y^{2}} & \leq & {z \leq \sqrt{18 - x^{2} - y^{2}}.} \end{array}$$

The first two ranges of variables describe a quarter disk in the first quadrant of the $xy$-plane. Hence the range for $\theta$ is $0 \leq \theta \leq \frac{\pi}{2}.$

The lower bound $z = \sqrt{x^{2} + y^{2}}$ is the upper half of a cone and the upper bound $z = \sqrt{18 - x^{2} - y^{2}}$ is the upper half of a sphere. Therefore, we have $0 \leq \rho \leq \sqrt{18},$ which is $0 \leq \rho \leq 3\sqrt{2}.$

For the ranges of $\varphi,$ we need to find where the cone and the sphere intersect, so solve the equation

$$\begin{array}{rll} {r^{2} + z^{2}} & = & 18 \\ {\left( \sqrt{x^{2} + y^{2}} \right)^{2} + z^{2}} & = & 18 \\ {z^{2} + z^{2}} & = & 18 \\ {2z^{2}} & = & 18 \\ z^{2} & = & 9 \\ z & = & 3. \end{array}$$

This gives

$$\begin{array}{rll} {3\sqrt{2}\ \text{cos}\ \varphi} & = & 3 \\ {\text{cos}\ \varphi} & = & \frac{1}{\sqrt{2}} \\ \varphi & = & {\frac{\pi}{4}.} \end{array}$$

Putting this together, we obtain

$${\int\limits_{y = 0}^{y = 3}\ {\int\limits_{x = 0}^{x = \sqrt{9 - y^{2}}}\ {\int\limits_{z = \sqrt{x^{2} + y^{2}}}^{z = \sqrt{18 - x^{2} - y^{2}}}{\left( {x^{2} + y^{2} + z^{2}} \right)dz\ dx\ dy}}}} = {\int\limits_{\varphi = 0}^{\varphi = \pi\text{/}4}\ {\int\limits_{\theta = 0}^{\theta = \pi\text{/}2}\ {\int\limits_{\rho = 0}^{\rho = 3\sqrt{2}}{\rho^{4}\text{sin}\ \varphi\ d\rho\ d\theta}}}}\ d\varphi.$$

Use rectangular, cylindrical, and spherical coordinates to set up triple integrals for finding the volume of the region inside the sphere $x^{2} + y^{2} + z^{2} = 4$ but outside the cylinder $x^{2} + y^{2} = 1.$

Now that we are familiar with the spherical coordinate system, let’s find the volume of some known geometric figures, such as spheres and ellipsoids.

Chapter Opener: Finding the Volume of l’Hemisphèric 章首问题：求 l’Hemisphèric 的体积

Find the volume of the spherical planetarium in l’Hemisphèric in Valencia, Spain, which is five stories tall and has a radius of approximately $50$ ft, using the equation $x^{2} + y^{2} + z^{2} = r^{2}.$

Solution 解

We calculate the volume of the ball in the first octant, where $x \geq 0,y \geq 0,$ and $z \geq 0,$ using spherical coordinates, and then multiply the result by $8$ for symmetry. Since we consider the region $D$ as the first octant in the integral, the ranges of the variables are

$$0 \leq \varphi \leq \frac{\pi}{2},0 \leq \rho \leq r,0 \leq \theta \leq \frac{\pi}{2}.$$

Therefore,

$$\begin{array}{cl} V & {= {\iiint\limits_{D}{dx\ dy\ dz = 8{\int\limits_{\theta = 0}^{\theta = \pi\text{/}2}\ {\int\limits_{\rho = 0}^{\rho = r}\ {\int\limits_{\varphi = 0}^{\varphi = \pi\text{/}2}{\rho^{2}\text{sin}\ \theta}}}}}}\ d\varphi\ d\rho\ d\theta} \\ & {= 8{\int\limits_{\varphi = 0}^{\varphi = \pi\text{/}2}{d\varphi{\int\limits_{\rho = 0}^{\rho = r}{\rho^{2}d\rho}}}}{\int\limits_{\theta = 0}^{\theta = \pi\text{/}2}{\text{sin}\ \theta\ d\theta}}} \\ & {= 8\left( \frac{\pi}{2} \right)\left( \frac{r^{3}}{3} \right)(1)} \\ & {= \frac{4}{3}\pi r^{3}.} \end{array}$$

This exactly matches with what we knew. So for a sphere with a radius of approximately $50$ ft, the volume is $\frac{4}{3}\pi{(50)}^{3} \approx 523,600{\ \text{ft}}^{3}.$

For the next example we find the volume of an ellipsoid.

Finding the Volume of an Ellipsoid 求椭球的体积

Find the volume of the ellipsoid $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1.$

Solution 解

We again use symmetry and evaluate the volume of the ellipsoid using spherical coordinates. As before, we use the first octant $x \geq 0,y \geq 0,$ and $z \geq 0$ and then multiply the result by $8.$

In this case the ranges of the variables are

$$0 \leq \varphi \leq \frac{\pi}{2},0,0 \leq \rho \leq 1,\ \text{and}\ 0 \leq \theta \leq \frac{\pi}{2}.$$

Also, we need to change the rectangular to spherical coordinates in this way:

$$x = a\rho\ \text{cos}\ \varphi\ \text{sin}\ \theta,y = b\rho\ \text{sin}\ \varphi\ \text{sin}\ \theta,\ \text{and}\ z = c\rho\ \text{cos}\ \theta.$$

Then the volume of the ellipsoid becomes

$$\begin{array}{cl} V & {= {\iiint\limits_{D}{dx\ dy\ dz}}} \\ & {= 8{\int\limits_{\theta = 0}^{\theta = \pi\text{/}2}\ {\int\limits_{\rho = 0}^{\rho = 1}\ {\int\limits_{\varphi = 0}^{\varphi = \pi\text{/}2}{abc\rho^{2}\text{sin}\ \theta}}}}\ d\varphi\ d\rho\ d\theta} \\ & {= 8abc{\int\limits_{\varphi = 0}^{\varphi = \pi\text{/}2}{d\varphi{\int\limits_{\rho = 0}^{\rho = 1}{\rho^{2}d\rho{\int\limits_{\theta = 0}^{\theta = \pi\text{/}2}{\text{sin}\ \theta\ d\theta}}}}}}} \\ & {= 8abc\left( \frac{\pi}{2} \right)\left( \frac{1}{3} \right)(1)} \\ & {= \frac{4}{3}\pi abc.} \end{array}$$

Finding the Volume of the Space Inside an Ellipsoid and Outside a Sphere 求椭球内、球外空间的体积

Find the volume of the space inside the ellipsoid $\frac{x^{2}}{75^{2}} + \frac{y^{2}}{80^{2}} + \frac{z^{2}}{90^{2}} = 1$ and outside the sphere $x^{2} + y^{2} + z^{2} = 50^{2}.$

Solution 解

This problem is directly related to the l’Hemisphèric structure. The volume of space inside the ellipsoid and outside the sphere might be useful to find the expense of heating or cooling that space. We can use the preceding two examples for the volume of the sphere and ellipsoid and then substract.

First we find the volume of the ellipsoid using $a = 75\ \text{ft,}$ $b = 80\ \text{ft,}$ and $c = 90\ \text{ft}$ in the result from Example 5.53. Hence the volume of the ellipsoid is

$$V_{\text{ellipsoid}} = \frac{4}{3}\pi(75)(80)(90) \approx 2,262,000{\ \text{ft}}^{3}.$$

From Example 5.52, the volume of the sphere is

$$V_{\text{sphere}} \approx 523,600{\ \text{ft}}^{3}.$$

Therefore, the volume of the space inside the ellipsoid $\frac{x^{2}}{75^{2}} + \frac{y^{2}}{80^{2}} + \frac{z^{2}}{90^{2}} = 1$ and outside the sphere $x^{2} + y^{2} + z^{2} = 50^{2}$ is approximately

$$V_{\text{Hemisferic}} = V_{\text{ellipsoid}} - V_{\text{sphere}} = 1,738,400{\ \text{ft}}^{3}.$$

Hot air balloons 热气球

Hot air ballooning is a relaxing, peaceful pastime that many people enjoy. Many balloonist gatherings take place around the world, such as the Albuquerque International Balloon Fiesta. The Albuquerque event is the largest hot air balloon festival in the world, with over $500$ balloons participating each year.

As the name implies, hot air balloons use hot air to generate lift. (Hot air is less dense than cooler air, so the balloon floats as long as the hot air stays hot.) The heat is generated by a propane burner suspended below the opening of the basket. Once the balloon takes off, the pilot controls the altitude of the balloon, either by using the burner to heat the air and ascend or by using a vent near the top of the balloon to release heated air and descend. The pilot has very little control over where the balloon goes, however—balloons are at the mercy of the winds. The uncertainty over where we will end up is one of the reasons balloonists are attracted to the sport.

In this project we use triple integrals to learn more about hot air balloons. We model the balloon in two pieces. The top of the balloon is modeled by a half sphere of radius $28$ feet. The bottom of the balloon is modeled by a frustum of a cone (think of an ice cream cone with the pointy end cut off). The radius of the large end of the frustum is $28$ feet and the radius of the small end of the frustum is $6$ feet. A graph of our balloon model and a cross-sectional diagram showing the dimensions are shown in the following figure.

We first want to find the volume of the balloon. If we look at the top part and the bottom part of the balloon separately, we see that they are geometric solids with known volume formulas. However, it is still worthwhile to set up and evaluate the integrals we would need to find the volume. If we calculate the volume using integration, we can use the known volume formulas to check our answers. This will help ensure that we have the integrals set up correctly for the later, more complicated stages of the project.

1. Find the volume of the balloon in two ways.

1. Use triple integrals to calculate the volume. Consider each part of the balloon separately. (Consider using spherical coordinates for the top part and cylindrical coordinates for the bottom part.)

2. Verify the answer using the formulas for the volume of a sphere, $V = \frac{4}{3}\pi r^{3},$ and for the volume of a cone, $V = \frac{1}{3}\pi r^{2}h.$

In reality, calculating the temperature at a point inside the balloon is a tremendously complicated endeavor. In fact, an entire branch of physics (thermodynamics) is devoted to studying heat and temperature. For the purposes of this project, however, we are going to make some simplifying assumptions about how temperature varies from point to point within the balloon. Assume that just prior to liftoff, the temperature (in degrees Fahrenheit) of the air inside the balloon varies according to the function

$$T_{0}(r,\theta,z) = \frac{z - r}{10} + 210.$$

2. What is the average temperature of the air in the balloon just prior to liftoff? (Again, look at each part of the balloon separately, and do not forget to convert the function into spherical coordinates when looking at the top part of the balloon.)

Now the pilot activates the burner for $10$ seconds. This action affects the temperature in a $12$-foot-wide column $20$ feet high, directly above the burner. A cross section of the balloon depicting this column in shown in the following figure.

Assume that after the pilot activates the burner for $10$ seconds, the temperature of the air in the column described above *increases* according to the formula

$$H(r,\theta,z) = -2z - 48.$$

Then the temperature of the air in the column is given by

$$T_{1}(r,\theta,z) = \frac{z - r}{10} + 210 + \left( {-2z - 48} \right),$$

while the temperature in the remainder of the balloon is still given by

$$T_{0}(r,\theta,z) = \frac{z - r}{10} + 210.$$

3. Find the average temperature of the air in the balloon after the pilot has activated the burner for $10$ seconds.

Section 5.5 Exercises 5.5 节习题

In the following exercises, evaluate the triple integrals $\iiint\limits_{B}{f(x,y,z)dV}$ over the solid $B.$

241.

$f(x,y,z) = z,$ $B = \left\{ \left( {x,y,z} \right) \middle| x^{2} + y^{2} \leq 9,x \geq 0,y \geq 0,0 \leq z \leq 1 \right\}$

242\.

$f(x,y,z) = xz^{2},$ $B = \left\{ \left( {x,y,z} \right) \middle| x^{2} + y^{2} \leq 16,x \geq 0,y \leq 0,-1 \leq z \leq 1 \right\}$

243.

$f(x,y,z) = xy,$ $B = \left\{ (x,y,z) \middle| x^{2} + y^{2} \leq 1,x \geq 0,y \geq 0,x \leq y,-1 \leq z \leq 1 \right\}$

244\.

$f(x,y,z) = x^{2} + y^{2},$ $B = \left\{ \left( {x,y,z} \right) \middle| x^{2} + y^{2} \leq 4,x \geq 0,x \leq y,0 \leq z \leq 3 \right\}$

245.

$f(x,y,z) = e^{\sqrt{x^{2} + y^{2}}},$ $B = \left\{ \left( {x,y,z} \right) \middle| 1 \leq x^{2} + y^{2} \leq 4,y \leq 0,x \leq y\sqrt{3},2 \leq z \leq 3 \right\}$

246\.

$f(x,y,z) = \sqrt{x^{2} + y^{2}},$ $B = \left\{ \left( {x,y,z} \right) \middle| 1 \leq x^{2} + y^{2} \leq 9,y \leq 0,0 \leq z \leq 1 \right\}$

247\.

1. Let $B$ be a cylindrical shell with inner radius $a,$ outer radius $b,$ and height $c,$ where $0 < a < b$ and $c > 0.$ Assume that a function $F$ defined on $B$ can be expressed in cylindrical coordinates as $F(x,y,z) = f(r) + h(z),$ where $f$ and $h$ are differentiable functions. If $\int\limits_{a}^{b}{\widetilde{f}(r)dr = 0}$ and $\widetilde{h}(0) = 0,$ where $\widetilde{f}$ and $\widetilde{h}$ are antiderivatives of $f$ and $h,$ respectively, show that

$${\iiint\limits_{B}{F\left( {x,y,z} \right)dV}} = 2\pi c\left( {b\widetilde{f}(b) - a\widetilde{f}(a)} \right) + \pi\left( {b^{2} - a^{2}} \right)\widetilde{h}(c).$$

2. Use the previous result to show that ${\iiint\limits_{B}\left( {z + \text{sin}\sqrt{x^{2} + y^{2}}} \right)}dx\ dy\ dz = 6\pi^{2}\left( {\pi - 2} \right),$ where $B$ is a cylindrical shell with inner radius $\pi,$ outer radius $2\pi,$ and height $2.$

248\.

1. Let $B$ be a cylindrical shell with inner radius $a,$ outer radius $b,$ and height $c,$ where $0 < a < b$ and $c > 0.$ Assume that a function $F$ defined on $B$ can be expressed in cylindrical coordinates as $F(x,y,z) = f(r)g(\theta)h(z),$ where $f,g,\ \text{and}\ h$ are differentiable functions. If ${\int\limits_{a}^{b}{\widetilde{f}(r)dr = 0}},$ where $\widetilde{f}$ is an antiderivative of $f,$ show that

$${\iiint\limits_{B}{F\left( {x,y,z} \right)dV}} = \left\lbrack {b\widetilde{f}(b) - a\widetilde{f}(a)} \right\rbrack\ \left\lbrack {\widetilde{g}(2\pi) - \widetilde{g}(0)} \right\rbrack\ \left\lbrack {\widetilde{h}(c) - \widetilde{h}(0)} \right\rbrack,$$

where $\widetilde{g}$ and $\widetilde{h}$ are antiderivatives of $g$ and $h,$ respectively.

2. Use the previous result to show that ${\iiint\limits_{B}{z\ \text{sin}\sqrt{x^{2} + y^{2}}}}dx\ dy\ dz = -12\pi^{2},$ where $B$ is a cylindrical shell with inner radius $\pi,$ outer radius $2\pi,$ and height $2.$

In the following exercises, the boundaries of the solid $E$ are given in cylindrical coordinates. Let $f~(r,\theta,z)$ be the corresponding function in cylindrical coordinates.

1. Define the region in cylindrical coordinates.

2. Convert the integral $\iiint\limits_{Ε}g(x,y,z)dV$ to cylindrical coordinates.

249.

$E$ is inside the right circular cylinder $r = 4\ \text{sin}\ \theta,$ above the $r\theta$-plane, and inside the sphere $r^{2} + z^{2} = 16.$

250\.

$E$ is inside the right circular cylinder $r = \text{cos}\ \theta,$ above the $r\theta$-plane, and inside the sphere $r^{2} + z^{2} = 9.$

251.

$E$ is located in the first octant and is bounded by the circular paraboloid $z = 9 - 3r^{2},$ the cylinder $r = \sqrt{3},$ and the plane $r\left( {\text{cos}\ \theta + \text{sin}\ \theta} \right) = 20 - z.$

252\.

$E$ is located in the first octant outside the circular paraboloid $z = 10 - 2r^{2}$ and inside the cylinder $r = \sqrt{5}$ and is bounded also by the planes $z = 20$ and $\theta = \frac{\pi}{4}.$

In the following exercises, the function $g$ and region $E$ are given in rectangular coordinates.

1. Express the region $Ε$ and the function $g$ in cylindrical coordinates. Let $f(r,\theta,z)$ be the corresponding function in cylindrical coordinates.

2. Convert the integral ${\iiint\limits_{E}{g\left( {x,y,z} \right)}}dV$ to cylindrical coordinates and evaluate it.

253.

$g\left( {x,y,z} \right) = \frac{1}{x + 3},$ $E = \left\{ \left( {x,y,z} \right) \middle| 0 \leq x^{2} + y^{2} \leq 9,x \geq 0,y \geq 0,0 \leq z \leq x + 3 \right\}$

254\.

$g\left( {x,y,z} \right) = x^{2} + y^{2},$ $E = \left\{ \left( {x,y,z} \right) \middle| 0 \leq x^{2} + y^{2} \leq 4,y \geq 0,0 \leq z \leq 3 - x \right\}$

255.

$g\left( {x,y,z} \right) = x,$ $E = \left\{ \left( {x,y,z} \right) \middle| 1 \leq y^{2} + z^{2} \leq 9,0 \leq x \leq 9 - y^{2} - z^{2} \right\}$

256\.

$g\left( {x,y,z} \right) = y,$ $E = \left\{ \left( {x,y,z} \right) \middle| 1 \leq x^{2} + z^{2} \leq 9,0 \leq y \leq 9 - x^{2} - z^{2} \right\}$

In the following exercises, find the volume of the solid $E$ whose boundaries are given in rectangular coordinates.

257.

$E$ is above the $xy$-plane, inside the cylinder $x^{2} + y^{2} = 1,$ and below the plane $z = 1.$

258\.

$E$ is below the plane $z = 1$ and inside the paraboloid $z = x^{2} + y^{2}.$

259.

$E$ is bounded by the circular cone $z = \sqrt{x^{2} + y^{2}}$ and $z = 1.$

260\.

$E$ is located above the $xy$-plane, below $z = 1,$ outside the one-sheeted hyperboloid $x^{2} + y^{2} - z^{2} = 1,$ and inside the cylinder $x^{2} + y^{2} = 2.$

261.

$E$ is located inside the cylinder $x^{2} + y^{2} = 1$ and between the circular paraboloids $z = 1 - x^{2} - y^{2}$ and $z = x^{2} + y^{2}.$

262\.

$E$ is located inside the sphere $x^{2} + y^{2} + z^{2} = 1,$ above the $xy$-plane, and inside the circular cone $z = \sqrt{x^{2} + y^{2}}.$

263.

$E$ is located inside the circular cone $x^{2} + y^{2} = \left( {z - 1} \right)^{2}$ and between the planes $z = 0$ and $z = 2.$

264\.

$E$ is located inside the cylinder $x^{2} + y^{2} = 1$, above the circular cone $z = 1 - \sqrt{x^{2} + y^{2}}$, and below the circular paraboloid $z = 1 + x^{2} + y^{2}$, and between the planes $z = 0$ and $z = 2$.

265.

\[T\] Use a computer algebra system (CAS) to graph the solid whose volume is given by the iterated integral in cylindrical coordinates ${\int\limits_{\text{−}{\pi\text{/}2}}^{\pi\text{/}2}\ {\int\limits_{0}^{1}\ {\int\limits_{r^{2}}^{r}{r\ dz\ dr\ d\theta}}}}.$ Find the volume $V$ of the solid. Round your answer to four decimal places.

266\.

\[T\] Use a CAS to graph the solid whose volume is given by the iterated integral in cylindrical coordinates ${\int\limits_{0}^{\pi\text{/}2}\ {\int\limits_{0}^{1}\ {\int\limits_{r^{4}}^{r}{r\ dz\ dr\ d\theta}}}}.$ Find the volume $V$ of the solid Round your answer to four decimal places.

267.

Convert the integral $\int\limits_{0}^{1}\ {\int\limits_{\text{−}\sqrt{1 - y^{2}}}^{\sqrt{1 - y^{2}}}\ {\int\limits_{x^{2} + y^{2}}^{\sqrt{x^{2} + y^{2}}}{xz\ dz\ dx\ dy}}}$ into an integral in cylindrical coordinates.

268\.

Convert the integral ${\int\limits_{0}^{2}\ {\int\limits_{0}^{y}\ {\int\limits_{0}^{1}\left( {xy + z} \right)}}}dz\ dx\ dy$ into an integral in cylindrical coordinates.

In the following exercises, evaluate the triple integral $\iiint\limits_{B}{f(x,y,z)dV}$ over the solid $B.$

269.

$f(x,y,z) = 1,$ $B = \left\{ \left( {x,y,z} \right) \middle| x^{2} + y^{2} + z^{2} \leq 90,z \geq 0 \right\}$

270\.

$f(x,y,z) = 1 - \sqrt{x^{2} + y^{2} + z^{2}},$ $B = \left\{ \left( {x,y,z} \right) \middle| x^{2} + y^{2} + z^{2} \leq 9,y \geq 0,z \geq 0 \right\}$

271.

$f(x,y,z) = \sqrt{x^{2} + y^{2}},$ $B$ is bounded above by the half-sphere $x^{2} + y^{2} + z^{2} = 9$ with $z \geq 0$ and below by the cone $z^{2} = x^{2} + y^{2}.$

272\.

$f(x,y,z) = z,$ $B$ is bounded above by the half-sphere $x^{2} + y^{2} + z^{2} = 16$ with $z \geq 0$ and below by the cone $z^{2} = x^{2} + y^{2}.$

273.

Show that if $F\left( {\rho,\theta,\varphi} \right) = f(\rho)g(\theta)h(\varphi)$ is a continuous function on the spherical box $B = \left\{ {\left. \left( {\rho,\theta,\varphi} \right) \right|a \leq \rho \leq b,\alpha \leq \theta \leq \beta,\gamma \leq \varphi \leq \psi} \right\},$ then

$${\iiint\limits_{B}{F\ dV}} = \left( {\int\limits_{a}^{b}{\rho^{2}f(\rho)dr}} \right)\left( {\int\limits_{\alpha}^{\beta}{g(\theta)d\theta}} \right)\left( {\int\limits_{\gamma}^{\psi}{h(\varphi)\text{sin}\ \varphi\ d\varphi}} \right).$$ 274.

1. A function $F$ is said to have spherical symmetry if it depends on the distance to the origin only, that is, it can be expressed in spherical coordinates as $F\left( {x,y,z} \right) = f(\rho),$ where $\rho = \sqrt{x^{2} + y^{2} + z^{2}}.$ Show that

$${\iiint\limits_{B}{F(x,y,z)dV}} = 2\pi{\int\limits_{a}^{b}{\rho^{2}f(\rho)d\rho}},$$

where $B$ is the region between the upper concentric hemispheres of radii $a$ and $b$ centered at the origin, with $0 < a < b$ and $F$ a spherical function defined on $B.$

2. Use the previous result to show that ${\iiint\limits_{B}{\left( {x^{2} + y^{2} + z^{2}} \right)\sqrt{x^{2} + y^{2} + z^{2}}\ dV}} = 21\pi,$ where

$$B = \left\{ \left( {x,y,z} \right) \middle| 1 \leq x^{2} + y^{2} + z^{2} \leq 2,z \geq 0 \right\}.$$

275\.

1. Let $B$ be the region between the upper concentric hemispheres of radii *a* and *b* centered at the origin and situated in the first octant, where $0 < a < b.$ Consider *F* a function defined on *B* whose form in spherical coordinates $\left( {\rho,\theta,\varphi} \right)$ is $F\left( {x,y,z} \right) = f(\rho)\text{cos}\ \varphi.$ Show that if $g(a) = g(b) = 0$ and ${\int\limits_{a}^{b}{h(\rho)d\rho = 0}},$ then

$${\iiint\limits_{B}{F(x,y,z)dV}} = \frac{\pi^{2}}{4}\left\lbrack {ah(a) - bh(b)} \right\rbrack,$$

where $g$ is an antiderivative of $f$ and $h$ is an antiderivative of $g.$

2. Use the previous result to show that ${\iiint\limits_{B}{\frac{z\ \text{cos}\sqrt{x^{2} + y^{2} + z^{2}}}{\sqrt{x^{2} + y^{2} + z^{2}}}dV}} = \frac{3\pi^{2}}{2},$ where $B$ is the region between the upper concentric hemispheres of radii $\pi$ and $2\pi$ centered at the origin and situated in the first octant.

In the following exercises, the function $g$ and region $E$ are given in rectangular coordinates.

1. Express the region $E$ and the function $g$ in spherical coordinates. Let $f(\rho,~\theta,~\varphi)~$ be the corresponding function in spherical coordinates

2. Convert the integral ${\iiint\limits_{E}{g\left( {x,y,z} \right)}}dV$ to spherical coordinates and evaluate it.

276\.

$g\left( {x,y,z} \right) = z;$ $E = \left\{ \left( {x,y,z} \right) \middle| 0 \leq x^{2} + y^{2} + z^{2} \leq 1,z \geq 0 \right\}$

277.

$g\left( {x,y,z} \right) = x + y;$ $E = \left\{ \left( {x,y,z} \right) \middle| 1 \leq x^{2} + y^{2} + z^{2} \leq 4,z \geq 0,y \geq 0 \right\}$

278\.

$g\left( {x,y,z} \right) = 2xy;$ $E = \left\{ \left( {x,y,z} \right) \middle| \sqrt{x^{2} + y^{2}} \leq z \leq \sqrt{1 - x^{2} - y^{2}},x \geq 0,y \geq 0 \right\}$

279.

$g\left( {x,y,z} \right) = z;$ $E = \left\{ \left( {x,y,z} \right) \middle| x^{2} + y^{2} + z^{2} - 2z \leq 0,\sqrt{x^{2} + y^{2}} \leq z \right\}$

In the following exercises, find the volume of the solid $E$ whose boundaries are given in rectangular coordinates.

280\.

$E = \left\{ \left( {x,y,z} \right) \middle| \sqrt{x^{2} + y^{2}} \leq z \leq \sqrt{16 - x^{2} - y^{2}},x \geq 0,y \geq 0 \right\}$

281.

$E = \left\{ \left( {x,y,z} \right) \middle| x^{2} + y^{2} + z^{2} - 2z \leq 0,\sqrt{x^{2} + y^{2}} \leq z \right\}$

282\.

Use spherical coordinates to find the volume of the solid situated inside the sphere $\rho = 1$ and outside the sphere $\rho = \text{cos}\ \varphi,$ with $\varphi \in \left\lbrack {0,\frac{\pi}{2}} \right\rbrack.$

283.

Use spherical coordinates to find the volume of the ball $\rho \leq 3$ that is situated between the cones $\varphi = \frac{\pi}{4}\ \text{and}\ \varphi = \frac{\pi}{3}.$

284\.

Convert the integral ${\int\limits_{-4}^{4}\ {\int\limits_{\text{−}\sqrt{16 - y^{2}}}^{\sqrt{16 - y^{2}}}\ {\int\limits_{\text{−}\sqrt{16 - x^{2} - y^{2}}}^{\sqrt{16 - x^{2} - y^{2}}}\left( {x^{2} + y^{2} + z^{2}} \right)}}}dz\ dx\ dy$ into an integral in spherical coordinates.

285.

Convert the integral ${\int\limits_{0}^{4}\ {\int\limits_{0}^{\sqrt{16 - x^{2}}}\ {\int\limits_{\text{−}\sqrt{16 - x^{2} - y^{2}}}^{\sqrt{16 - x^{2} - y^{2}}}\left( {x^{2} + y^{2} + z^{2}} \right)^{2}}}}dz\ dy\ dx$ into an integral in spherical coordinates.

286\.

Convert the integral $\int\limits_{-2\sqrt{2}}^{2\sqrt{2}}\ \int\limits_{- \sqrt{8 - x^{2}}}^{\sqrt{8 - x^{2}}}\ \int\limits_{\sqrt{x^{2} + y^{2}}}^{\sqrt{16 - x^{2} - y^{2}}}dz\ dy\ dx$ into an integral in spherical coordinates and evaluate it.

287.

\[T\] Use a CAS to graph the solid whose volume is given by the iterated integral in spherical coordinates ${\int\limits_{\pi\text{/}2}^{\pi}\ {\int\limits_{{5\pi}\text{/}6}^{\pi\text{/}6}\ {\int\limits_{0}^{2}{\rho^{2}\text{sin}\ \varphi}}}}\ d\rho\ d\varphi\ d\theta.$ Find the volume $V$ of the solid. Round your answer to three decimal places.

288\.

\[T\] Use a CAS to graph the solid whose volume is given by the iterated integral in spherical coordinates as ${\int\limits_{0}^{2\pi}\ {\int\limits_{\pi\text{/}4}^{3\pi\text{/}4}\ {\int\limits_{0}^{1}{\rho^{2}\text{sin}\ \varphi}}}}\ d\rho\ d\varphi\ d\theta.$ Find the volume $V$ of the solid. Round your answer to three decimal places.

289.

\[T\] Use a CAS to evaluate the integral ${\iiint\limits_{E}\left( {x^{2} + y^{2}} \right)}dV$ where $E$ lies above the paraboloid $z = x^{2} + y^{2}$ and below the plane $z = 3y.$

290\.

\[T\]

1. Evaluate the integral ${\iiint\limits_{E}{e^{\sqrt{x^{2} + y^{2} + z^{2}}}dV}},$ where $E$ is bounded by the spheres $4x^{2} + 4y^{2} + 4z^{2} = 1$ and $x^{2} + y^{2} + z^{2} = 1.$

2. Use a CAS to find an approximation of the previous integral. Round your answer to two decimal places.

291.

Express the volume of the solid inside the sphere $x^{2} + y^{2} + z^{2} = 16$ and outside the cylinder $x^{2} + y^{2} = 4$ as triple integrals in cylindrical coordinates and spherical coordinates.

292\.

Express the volume of the solid inside the sphere $x^{2} + y^{2} + z^{2} = 16$ and outside the cylinder $x^{2} + y^{2} = 4$ that is located in the first octant as triple integrals in cylindrical coordinates and spherical coordinates.

293.

The power emitted by an antenna has a power density per unit volume given in spherical coordinates by

$p\left( {\rho,\theta,\varphi} \right) = \frac{P_{0}}{\rho^{2}}\text{cos}^{2}\theta\ \text{sin}^{4}\varphi,$where $P_{0}$ is a constant with units in watts. The total power within a sphere $B$ of radius $r$ meters is defined as $P = {\iiint\limits_{B}{p\left( {\rho,\theta,\varphi} \right)}}dV.$ Find the total power $P$ within a sphere of radius 20 meters.

294\.

Use the preceding exercise to find the total power within a sphere $B$ of radius 5 meters when the power density per unit volume is given by $p\left( {\rho,\theta,\varphi} \right) = \frac{30}{\rho^{2}}\text{cos}^{2}\theta\ \text{sin}^{4}\varphi.$

295.

A charge cloud contained in a sphere $B$ of radius *r* centimeters centered at the origin has its charge density given by $q\left( {x,y,z} \right) = k\sqrt{x^{2} + y^{2} + z^{2}}\frac{\mu\ \text{C}}{\text{cm}^{3}},$ where $k > 0.$ The total charge contained in $B$ is given by $Q = {\iiint\limits_{B}{q\left( {x,y,z} \right)}}dV.$ Find the total charge $Q.$

296\.

Use the preceding exercise to find the total charge cloud contained in the unit sphere if the charge density is $q\left( {x,y,z} \right) = 20\sqrt{x^{2} + y^{2} + z^{2}}\frac{\mu\ \text{C}}{\text{cm}^{3}}.$

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5.6 Calculating Centers of Mass and Moments of Inertia 5.6 质心与转动惯量的计算

• 5.6.1 Use double integrals to locate the center of mass of a two-dimensional object.
• 5.6.2 Use double integrals to find the moment of inertia of a two-dimensional object.
• 5.6.3 Use triple integrals to locate the center of mass of a three-dimensional object.

• 5.6.1 用二重积分确定二维物体的质心。
• 5.6.2 用二重积分求二维物体的转动惯量。
• 5.6.3 用三重积分确定三维物体的质心。

We have already discussed a few applications of multiple integrals, such as finding areas, volumes, and the average value of a function over a bounded region. In this section we develop computational techniques for finding the center of mass and moments of inertia of several types of physical objects, using double integrals for a lamina (flat plate) and triple integrals for a three-dimensional object with variable density. The density is usually considered to be a constant number when the lamina or the object is homogeneous; that is, the object has uniform density.

Center of Mass in Two Dimensions 二维平面上的质心

The center of mass is also known as the center of gravity if the object is in a uniform gravitational field. If the object has uniform density, the center of mass is the geometric center of the object, which is called the centroid. Figure 5.64 shows a point $P$ as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.

To find the coordinates of the center of mass $P(\overset{\text{−}}{x},\overset{\text{−}}{y})$ of a lamina, we need to find the moment $M_{x}$ of the lamina about the $x\text{-axis}$ and the moment $M_{y}$ about the $y\text{-axis}\text{.}$ We also need to find the mass $m$ of the lamina. Then

$$\overset{\text{−}}{x} = \frac{M_{y}}{m}\ \text{and}\ \overset{\text{−}}{y} = \frac{M_{x}}{m}.$$

Refer to Moments and Centers of Mass for the definitions and the methods of single integration to find the center of mass of a one-dimensional object (for example, a thin rod). We are going to use a similar idea here except that the object is a two-dimensional lamina and we use a double integral.

If we allow a constant density function, then $\overset{\text{−}}{x} = \frac{M_{y}}{m}\ \text{and}\ \overset{\text{−}}{y} = \frac{M_{x}}{m}$ give the *centroid* of the lamina.

Suppose that the lamina occupies a region $R$ in the $xy\text{-plane},$ and let $\rho\left( {x,y} \right)$ be its density (in units of mass per unit area) at any point $\left( {x,y} \right).$ Hence, $\rho\left( {x,y} \right) = \underset{\text{Δ}A\rightarrow 0}{\text{lim}}\frac{\text{Δ}m}{\text{Δ}A},$ where $\text{Δ}m$ and $\text{Δ}A$ are the mass and area of a small rectangle containing the point $\left( {x,y} \right)$ and the limit is taken as the dimensions of the rectangle go to $0$ (see the following figure).

Just as before, we divide the region $R$ into tiny rectangles $R_{ij}$ with area $\text{Δ}A$ and choose $\left( {x_{ij}^{*},y_{ij}^{*}} \right)$ as sample points. Then the mass $m_{ij}$ of each $R_{ij}$ is equal to $\rho\left( {x_{ij}^{*},y_{ij}^{*} \right)\text{Δ}A$ (Figure 5.66). Let $k$ and $l$ be the number of subintervals in $x$ and $y,$ respectively. Also, note that the shape might not always be rectangular but the limit works anyway, as seen in previous sections.

Hence, the mass of the lamina is

$$m = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}m_{ij}}} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}{\rho(x_{ij}^{*},y_{ij}^{*})\text{Δ}A = {\iint\limits_{R}{\rho(x,y)dA}}}}}.$$ (5.13)

Let’s see an example now of finding the total mass of a triangular lamina.

Finding the Total Mass of a Lamina 求薄板的总质量

Consider a triangular lamina $R$ with vertices $\left( {0,0} \right),\left( {0,3} \right),$ $\left( {3,0} \right)$ and with density $\rho\left( {x,y} \right) = xy{\ \text{kg/m}}^{2}.$ Find the total mass.

Solution 解

A sketch of the region $R$ is always helpful, as shown in the following figure.

Using the expression developed for mass, we see that

$$\begin{array}{cl} m & {= {\iint\limits_{R}{dm = {\iint\limits_{R}{\rho\left( {x,y} \right)}}}}dA = {\int\limits_{x = 0}^{x = 3}\ {\int\limits_{y = 0}^{y = 3 - x}{xy\ dy\ dx = {\int\limits_{x = 0}^{x = 3}\left\lbrack \left. {x\frac{y^{2}}{2}} \right|_{y = 0}^{y = 3 - x} \right\rbrack}}}}dx} \\ & {= {\int\limits_{x = 0}^{x = 3}\frac{1}{2}}x\left( {3 - x} \right)^{2}dx = \left. \left\lbrack {\frac{9x^{2}}{4} - x^{3} + \frac{x^{4}}{8}} \right\rbrack \right|_{x = 0}^{x = 3}} \\ & {= \frac{27}{8}.} \end{array}$$

The computation is straightforward, giving the answer $m = \frac{27}{8}\ \text{kg}\text{.}$

Consider the same region $R$ as in the previous example, and use the density function $\rho\left( {x,y} \right) = \sqrt{xy}.$ Find the total mass. *Hint:* Use trigonometric substitution $\sqrt{x} = \sqrt{3}\sin\theta$ and then use the power reducing formulas for trigonometric functions.

Now that we have established the expression for mass, we have the tools we need for calculating moments and centers of mass. The moment $M_{x}$ about the $x\text{-axis}$ for $R$ is the limit of the sums of moments of the regions $R_{ij}$ about the $x\text{-axis}\text{.}$ Hence

$$M_{x} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}\left( y_{{}_{ij}}^{*} \right)}}m_{ij} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}\left( y_{{}_{ij}}^{*} \right)}}\rho\left( {x_{{}_{ij}}^{*},y_{{}_{ij}}^{*}} \right)\text{Δ}A = {\iint\limits_{R}{y\rho\left( {x,y} \right)dA}}.$$ (5.14)

Similarly, the moment $M_{y}$ about the $y\text{-axis}$ for $R$ is the limit of the sums of moments of the regions $R_{ij}$ about the $y\text{-axis}\text{.}$ Hence

$$M_{y} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}\left( x_{{}_{ij}}^{*} \right)}}m_{ij} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}\left( y_{{}_{ij}}^{*} \right)}}\rho\left( {x_{{}_{ij}}^{*},y_{{}_{ij}}^{*}} \right)\text{Δ}A = {\iint\limits_{R}{x\rho\left( {x,y} \right)dA}}.$$ (5.15)

Finding Moments 求矩

Consider the same triangular lamina $R$ with vertices ${\left( {0,0} \right),\left( {0,3} \right),}\ \left( {3,0} \right)$ and with density $\rho\left( {x,y} \right) = xy.$ Find the moments $M_{x}$ and $M_{y}.$

Solution 解

Use double integrals for each moment and compute their values:

$$M_{x} = {\iint\limits_{R}{y\rho\left( {x,y} \right)dA = {\int\limits_{x = 0}^{x = 3}\ {\int\limits_{y = 0}^{y = 3 - x}{xy^{2}\ dy\ dx}}}}} = \frac{81}{20},$$ $$M_{y} = {\iint\limits_{R}{x\rho\left( {x,y} \right)dA = {\int\limits_{x = 0}^{x = 3}\ {\int\limits_{y = 0}^{y = 3 - x}{x^{2}\ yd\ y\ dx}}}}} = \frac{81}{20}.$$

The computation is quite straightforward.

Consider the same lamina $R$ as above, and use the density function $\rho\left( {x,y} \right) = \sqrt{xy}.$ Find the moments $M_{x}$ and $M_{y}.$

Finally we are ready to restate the expressions for the center of mass in terms of integrals. We denote the *x*-coordinate of the center of mass by $\overset{\text{−}}{x}$ and the *y*-coordinate by $\overset{\text{−}}{y}.$ Specifically,

$$\overset{\text{−}}{x} = \frac{M_{y}}{m} = \frac{\iint\limits_{R}{x\rho\left( {x,y} \right)dA}}{\iint\limits_{R}{\rho\left( {x,y} \right)dA}}\mspace{7mu}\text{and}\mspace{7mu}\overset{\text{−}}{y} = \frac{M_{x}}{m}\ = \frac{\iint\limits_{R}{y\rho\left( {x,y} \right)dA}}{\iint\limits_{R}{\rho\left( {x,y} \right)dA}}.$$ (5.16)

Finding the Center of Mass 求质心

Again consider the same triangular region $R$ with vertices $\left( {0,0} \right),\left( {0,3} \right),$ $\left( {3,0} \right)$ and with density function $\rho\left( {x,y} \right) = xy.$ Find the center of mass.

Solution 解

Using the formulas we developed, we have

$$\overset{\text{−}}{x} = \frac{M_{y}}{m} = \frac{\iint\limits_{R}{x\rho\left( {x,y} \right)dA}}{\iint\limits_{R}{\rho\left( {x,y} \right)dA}} = \frac{81\text{/}20}{27\text{/}8} = \frac{6}{5},$$ $$\overset{\text{−}}{y} = \frac{M_{x}}{m} = \frac{\iint\limits_{R}{y\rho\left( {x,y} \right)dA}}{\iint\limits_{R}{\rho\left( {x,y} \right)dA}} = \frac{81\text{/}20}{27\text{/}8} = \frac{6}{5}.$$

Therefore, the center of mass is the point $\left( {\frac{6}{5},\frac{6}{5}} \right).$

Analysis 分析

If we choose the density $\rho\left( {x,y} \right)$ instead to be uniform throughout the region (i.e., constant), such as the value 1 (any constant will do), then we can compute the centroid,

$$\begin{array}{l} \\ \\ \\ \\ {x_{c} = \frac{M_{y}}{m} = \frac{\iint\limits_{R}{x\ dA}}{\iint\limits_{R}{dA}} = \frac{9\text{/}2}{9\text{/}2} = 1,} \\ {y_{c} = \frac{M_{x}}{m}\ = \frac{\iint\limits_{R}{y\ dA}}{\iint\limits_{R}{dA}} = \frac{9\text{/}2}{9\text{/}2} = 1.} \end{array}$$

Notice that the center of mass $\left( {\frac{6}{5},\frac{6}{5}} \right)$ is not exactly the same as the centroid $\left( {1,1} \right)$ of the triangular region. This is due to the variable density of $R.$ If the density is constant, then we just use $\rho\left( {x,y} \right) = c$ (constant). This value cancels out from the formulas, so for a constant density, the center of mass coincides with the centroid of the lamina.

Again use the same region $R$ as above and the density function $\rho\left( {x,y} \right) = \sqrt{xy}.$ Find the center of mass.

Once again, based on the comments at the end of Example 5.57, we have expressions for the centroid of a region on the plane:

$$x_{c} = \frac{M_{y}}{m} = \frac{\iint\limits_{R}{x\ dA}}{\iint\limits_{R}{dA}}\ \text{and}\ y_{c} = \frac{M_{x}}{m}\ \frac{\iint\limits_{R}{y\ dA}}{\iint\limits_{R}{dA}}.$$

We should use these formulas and verify the centroid of the triangular region $R$ referred to in the last three examples.

Finding Mass, Moments, and Center of Mass 求质量、矩与质心

Find the mass, moments, and the center of mass of the lamina of density $\rho\left( {x,y} \right) = x + y$ occupying the region $R$ under the curve $y = x^{2}$ in the interval $0 \leq x \leq 2$ (see the following figure).

Solution 解

First we compute the mass $m.$ We need to describe the region between the graph of $y = x^{2}$ and the vertical lines $x = 0$ and $x = 2\text{:}$

$$\begin{array}{cl} m & {= {\iint\limits_{R}{dm}} = {\iint\limits_{R}{\rho\left( {x,y} \right)dA = {\int\limits_{x = 0}^{x = 2}\ {\int\limits_{y = 0}^{y = x^{2}}{\left( {x + y} \right)dy\ dx = {\int\limits_{x = 0}^{x = 2}\left\lbrack \left. {xy + \frac{y^{2}}{2}} \right|_{y = 0}^{y = x^{2}} \right\rbrack}}}}}}dx} \\ & {= {\int\limits_{x = 0}^{x = 2}\left\lbrack {x^{3} + \frac{x^{4}}{2}} \right\rbrack}dx = \left. \left\lbrack {\frac{x^{4}}{4} + \frac{x^{5}}{10}} \right\rbrack \right|_{x = 0}^{x = 2} = \frac{36}{5}.} \end{array}$$

Now compute the moments $M_{x}$ and $M_{y}\text{:}$

$$M_{x} = {\iint\limits_{R}{y\rho\left( {x,y} \right)}}dA = {\int\limits_{x = 0}^{x = 2}\ {\int\limits_{y = 0}^{y = x^{2}}{y\left( {x + y} \right)dy\ dx = \frac{80}{7}}}},$$ $$M_{y} = {\iint\limits_{R}{x\rho\left( {x,y} \right)}}dA = {\int\limits_{x = 0}^{x = 2}\ {\int\limits_{y = 0}^{y = x^{2}}{x\left( {x + y} \right)dy\ dx = \frac{176}{15}}}}.$$

Finally, evaluate the center of mass,

$$\begin{array}{l} \\ \\ \\ \\ {\overset{\text{−}}{x} = \frac{M_{y}}{m} = \frac{\iint\limits_{R}{x\rho\left( {x,y} \right)dA}}{\iint\limits_{R}{\rho\left( {x,y} \right)dA}} = \frac{176\text{/}15}{36\text{/}5} = \frac{44}{27},} \\ {\overset{\text{−}}{y} = \frac{M_{x}}{m} = \frac{\iint\limits_{R}{y\rho\left( {x,y} \right)dA}}{\iint\limits_{R}{\rho\left( {x,y} \right)dA}} = \frac{80\text{/}7}{36\text{/}5} = \frac{100}{63}.} \end{array}$$

Hence the center of mass is $(\overset{\text{−}}{x},\overset{\text{−}}{y}) = \left( {\frac{44}{27},\frac{100}{63}} \right).$

Calculate the mass, moments, and the center of mass of the region between the curves $y = x$ and $y = x^{2}$ with the density function $\rho\left( {x,y} \right) = x$ in the interval $0 \leq x \leq 1.$

Finding a Centroid 求质心

Find the centroid of the region under the curve $y = e^{x}$ over the interval $1 \leq x \leq 3$ (see the following figure).

Solution 解

To compute the centroid, we assume that the density function is constant and hence it cancels out:

$$\begin{array}{l} \\ \\ \\ \\ \\ {x_{c} = \frac{M_{y}}{m} = \frac{\iint\limits_{R}{x\ dA}}{\iint\limits_{R}{dA}}\ \text{and}\ y_{c} = \frac{M_{x}}{m} = \frac{\iint\limits_{R}{y\ dA}}{\iint\limits_{R}{dA}},} \\ {x_{c} = \frac{M_{y}}{m} = \frac{\iint\limits_{R}{x\ dA}}{\iint\limits_{R}{dA}} = \frac{\int\limits_{x = 1}^{x = 3}\ {\int\limits_{y = 0}^{y = e^{x}}{x\ dy\ dx}}}{\int\limits_{x = 1}^{x = 3}\ {\int\limits_{y = 0}^{y = e^{x}}{dy\ dx}}} = \frac{\int\limits_{x = 1}^{x = 3}{xe^{x}dx}}{\int\limits_{x = 1}^{x = 3}{e^{x}dx}} = \frac{2e^{3}}{e^{3} - e} = \frac{2e^{2}}{e^{2} - 1},} \\ {y_{c} = \frac{M_{x}}{m} = \frac{\iint\limits_{R}{y\ dA}}{\iint\limits_{R}{dA}} = \frac{\int\limits_{x = 1}^{x = 3}\ {\int\limits_{y = 0}^{y = e^{x}}{y\ dy\ dx}}}{\int\limits_{x = 1}^{x = 3}\ {\int\limits_{y = 0}^{y = e^{x}}{dy\ dx}}} = \frac{\int\limits_{x = 1}^{x = 3}{\frac{e^{2x}}{2}dx}}{\int\limits_{x = 1}^{x = 3}{e^{x}dx}} = \frac{\frac{1}{4}e^{2}\left( {e^{4} - 1} \right)}{e\left( {e^{2} - 1} \right)} = \frac{1}{4}e\left( {e^{2} + 1} \right).} \end{array}$$

Thus the centroid of the region is

$$\left( {x_{c},y_{c}} \right) = \left( {\frac{2e^{2}}{e^{2} - 1},\frac{1}{4}e\left( {e^{2} + 1} \right)} \right).$$

Calculate the centroid of the region between the curves $y = x$ and $y = \sqrt{x}$ with uniform density in the interval $0 \leq x \leq 1.$

Moments of Inertia 转动惯量

For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition of moments and centers of mass in Section 6.6 of Volume 1. The moment of inertia of a particle of mass $m$ about an axis is $mr^{2},$ where $r$ is the distance of the particle from the axis. We can see from Figure 5.66 that the moment of inertia of the subrectangle $R_{ij}$ about the $x\text{-axis}$ is ${(y_{ij}^{*})}^{2}\rho(x_{ij}^{*},y_{ij}^{*})\text{Δ}A.$ Similarly, the moment of inertia of the subrectangle $R_{ij}$ about the $y\text{-axis}$ is ${(x_{ij}^{*})}^{2}\rho(x_{ij}^{*},y_{ij}^{*})\text{Δ}A.$ The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis.

The moment of inertia $I_{x}$ about the $x\text{-axis}$ for the region $R$ is the limit of the sum of moments of inertia of the regions $R_{ij}$ about the $x\text{-axis}\text{.}$ Hence

$$I_{x} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}\left( y_{ij}^{*} \right)}}^{2}m_{ij} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}\left( y_{ij}^{*} \right)}}^{2}\rho\left( {x_{ij}^{*},y_{ij}^{*}} \right)\text{Δ}A = {\iint\limits_{R}{y^{2}\rho\left( {x,y} \right)}}dA.$$

Similarly, the moment of inertia $I_{y}$ about the $y\text{-axis}$ for $R$ is the limit of the sum of moments of inertia of the regions $R_{ij}$ about the $y\text{-axis}\text{.}$ Hence

$$I_{y} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}\left( x_{ij}^{*} \right)}}^{2}m_{ij} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}\left( x_{ij}^{*} \right)}}^{2}\rho\left( {x_{ij}^{*},y_{ij}^{*}} \right)\text{Δ}A = {\iint\limits_{R}{x^{2}\rho\left( {x,y} \right)}}dA.$$

Sometimes, we need to find the moment of inertia of an object about the origin, which is known as the polar moment of inertia. We denote this by $I_{0}$ and obtain it by adding the moments of inertia $I_{x}$ and $I_{y}.$ Hence

$$I_{0} = I_{x} + I_{y} = {\iint\limits_{R}\left( {x^{2} + y^{2}} \right)}\rho\left( {x,y} \right)dA.$$

All these expressions can be written in polar coordinates by substituting $x = r\ \text{cos}\ \theta,$ $y = r\ \text{sin}\ \theta,$ and $dA = r\ dr\ d\theta.$ For example, $I_{0} = {\iint\limits_{R}{r^{2}\rho\left( {r\ \text{cos}\ \theta,r\ \text{sin}\ \theta} \right)}}dA.$

Finding Moments of Inertia for a Triangular Lamina 求三角形薄板的转动惯量

Use the triangular region $R$ with vertices $\left( {0,0} \right),\left( {2,2} \right),$ and $\left( {2,0} \right)$ and with density $\rho\left( {x,y} \right) = xy$ as in previous examples. Find the moments of inertia.

Solution 解

Using the expressions established above for the moments of inertia, we have

$$\begin{array}{cll} I_{x} & = & {{\iint\limits_{R}{y^{2}\rho\left( {x,y} \right)}}dA = {\int\limits_{x = 0}^{x = 2}\ {\int\limits_{y = 0}^{y = x}{xy^{3}dy\ dx}}} = \frac{8}{3},} \\ I_{y} & = & {{\iint\limits_{R}{x^{2}\rho\left( {x,y} \right)}}dA = {\int\limits_{x = 0}^{x = 2}\ {\int\limits_{y = 0}^{y = x}{x^{3}y\ dy\ dx}}} = \frac{16}{3},} \\ I_{0} & = & {{\iint\limits_{R}\left( {x^{2} + y^{2}} \right)}\rho(x,y)dA = {\int\limits_{0}^{2}\ {\int\limits_{0}^{x}{\left( {x^{2} + y^{2}} \right)xy\ dy\ dx}}}} \\ & = & {I_{x} + I_{y} = 8.} \end{array}$$

Again use the same region $R$ as above and the density function $\rho\left( {x,y} \right) = \sqrt{xy}.$ Find the moments of inertia.

Solution 解

If we compute the mass of this region we find that $m = 2.$ We found the moments of inertia of this lamina in Example 5.58. From these data, the radii of gyration with respect to the $x\text{-axis,}$ $y\text{-axis,}$ and the origin are, respectively,

$$\begin{array}{rll} R_{x} & = & {\sqrt{\frac{I_{x}}{m}} = \sqrt{\frac{8\text{/}3}{2}} = \sqrt{\frac{8}{6}} = \frac{2\sqrt{3}}{3},} \\ R_{y} & = & {\sqrt{\frac{I_{y}}{m}} = \sqrt{\frac{16\text{/}3}{2}} = \sqrt{\frac{8}{3}} = \frac{2\sqrt{6}}{3},} \\ R_{0} & = & {\sqrt{\frac{I_{0}}{m}} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2.} \end{array}$$

Use the same region $R$ from Example 5.61 and the density function $\rho\left( {x,y} \right) = \sqrt{xy}.$ Find the radii of gyration with respect to the $x\text{-axis,}$ the $y\text{-axis,}$ and the origin.

Center of Mass and Moments of Inertia in Three Dimensions 三维空间中的质心与转动惯量

All the expressions of double integrals discussed so far can be modified to become triple integrals.

If we have a solid object $Q$ with a density function $\rho\left( {x,y,z} \right)$ at any point $\left( {x,y,z} \right)$ in space, then its mass is

$$m = {\iiint\limits_{Q}{\rho\left( {x,y,z} \right)}}dV.$$

Its moments about the $xy\text{-plane,}$ the $xz\text{-plane,}$ and the $yz\text{-plane}$ are

$$\begin{array}{l} {{M_{xy} = {\iiint\limits_{Q}{z\rho\left( {x,y,z} \right)}}dV,}\ {M_{xz} = {\iiint\limits_{Q}{y\rho\left( {x,y,z} \right)}}dV,}} \\ {M_{yz} = {\iiint\limits_{Q}{x\rho\left( {x,y,z} \right)}}dV.} \end{array}$$

If the center of mass of the object is the point $\left( {\overset{\text{−}}{x},\overset{\text{−}}{y},\overset{\text{−}}{z}} \right),$ then

$$\overset{\text{−}}{x} = \frac{M_{yz}}{m},\ \overset{\text{−}}{y} = \frac{M_{xz}}{m},\overset{\text{−}}{z} = \frac{M_{xy}}{m}.$$

Also, if the solid object is homogeneous (with constant density), then the center of mass becomes the centroid of the solid. Finally, the moments of inertia about the $yz\text{-plane,}$ the $xz\text{-plane,}$ and the $xy\text{-plane}$ are

$$\begin{array}{l} \\ {I_{x} = {\iiint\limits_{Q}\left( {y^{2} + z^{2}} \right)}\rho\left( {x,y,z} \right)dV,} \\ {I_{y} = {\iiint\limits_{Q}\left( {x^{2} + z^{2}} \right)}\rho\left( {x,y,z} \right)dV,} \\ {I_{z} = {\iiint\limits_{Q}\left( {x^{2} + y^{2}} \right)}\rho\left( {x,y,z} \right)dV.} \end{array}$$

Finding the Mass of a Solid 求立体的质量

Suppose that $Q$ is a solid region bounded by $x + 2y + 3z = 6$ and the coordinate planes and has density $\rho\left( {x,y,z} \right) = x^{2}yz.$ Find the total mass.

Solution 解

The region $Q$ is a tetrahedron (Figure 5.70) meeting the axes at the points $\left( {6,0,0} \right),\left( {0,3,0} \right),$ and $\left( {0,0,2} \right).$ To find the limits of integration, let $z = 0$ in the slanted plane $z = \frac{1}{3}\left( {6 - x - 2y} \right).$ Then for $x$ and $y$ find the projection of $Q$ onto the $xy\text{-plane,}$ which is bounded by the axes and the line $x + 2y = 6.$ Hence the mass is

$$m = {\iiint\limits_{Q}{\rho\left( {x,y,z} \right)dV = {\int\limits_{x = 0}^{x = 6}\ {\int\limits_{y = 0}^{y = 1\text{/}2{({6 - x})}}\ {\int\limits_{z = 0}^{z = 1\text{/}3{({6 - x - 2y})}}{x^{2}yz\ dz\ dy\ dx = \frac{108}{35} \approx 3.086}}}}}}.$$

Consider the same region $Q$ (Figure 5.70), and use the density function $\rho\left( {x,y,z} \right) = xy^{2}z.$ Find the mass.

Finding the Center of Mass of a Solid 求立体的质心

Suppose $Q$ is a solid region bounded by the plane $x + 2y + 3z = 6$ and the coordinate planes with density $\rho\left( {x,y,z} \right) = x^{2}yz$ (see Figure 5.70). Find the center of mass using decimal approximation. Use the mass found in Example 5.62

Solution 解

We have used this tetrahedron before and know the limits of integration, so we can proceed to the computations right away. First, we need to find the moments about the $xy\text{-plane,}$ the $xz\text{-plane,}$ and the $yz\text{-plane:}$

$$\begin{array}{l} \\ \\ \\ {M_{xy} = {\iiint\limits_{Q}{z\rho\left( {x,y,z} \right)}}dV = {\int\limits_{x = 0}^{x = 6}\ {\int\limits_{y = 0}^{y = 1\text{/}2{({6 - x})}}\ {\int\limits_{z = 0}^{z = 1\text{/}3{({6 - x - 2y})}}{x^{2}yz^{2}dz\ dy\ dx = \frac{54}{35}}}}} \approx 1.543,} \\ {M_{xz} = {\iiint\limits_{Q}{y\rho\left( {x,y,z} \right)}}dV = {\int\limits_{x = 0}^{x = 6}\ {\int\limits_{y = 0}^{y = 1\text{/}2{({6 - x})}}\ {\int\limits_{z = 0}^{z = 1\text{/}3{({6 - x - 2y})}}{x^{2}y^{2}z\ dz\ dy\ dx = \frac{81}{35}}}}} \approx 2.314,} \\ {M_{yz} = {\iiint\limits_{Q}{x\rho\left( {x,y,z} \right)}}dV = {\int\limits_{x = 0}^{x = 6}\ {\int\limits_{y = 0}^{y = 1\text{/}2{({6 - x})}}\ {\int\limits_{z = 0}^{z = 1\text{/}3{({6 - x - 2y})}}{x^{3}yz\ dz\ dy\ dx = \frac{243}{35}}}}} \approx 6.943.} \end{array}$$

Hence the center of mass is

$$\begin{array}{l} \\ \\ \\ {\overset{\text{−}}{x} = \frac{M_{yz}}{m},\overset{\text{−}}{y} = \frac{M_{xz}}{m},\overset{\text{−}}{z} = \frac{M_{xy}}{m},} \\ {\overset{\text{−}}{x} = \frac{M_{yz}}{m} = \frac{243\text{/}35}{108\text{/}35} = \frac{243}{108} = 2.25,} \\ {\overset{\text{−}}{y} = \frac{M_{xz}}{m} = \frac{81\text{/}35}{108\text{/}35} = \frac{81}{108} = 0.75,} \\ {\overset{\text{−}}{z} = \frac{M_{xy}}{m} = \frac{54\text{/}35}{108\text{/}35} = \frac{54}{108} = 0.5.} \end{array}$$

The center of mass for the tetrahedron $Q$ is the point $\left( {2.25,0.75,0.5} \right).$

Consider the same region $Q$ (Figure 5.70) and use the density function $\rho\left( {x,y,z} \right) = xy^{2}z.$ Find the center of mass.

We conclude this section with an example of finding moments of inertia $I_{x},I_{y},$ and $I_{z}.$

Finding the Moments of Inertia of a Solid 求立体的转动惯量

Suppose that $Q$ is a solid region and is bounded by $x + 2y + 3z = 6$ and the coordinate planes with density $\rho\left( {x,y,z} \right) = x^{2}yz$ (see Figure 5.70). Find the moments of inertia of the tetrahedron $Q$ about the $yz\text{-plane,}$ the $xz\text{-plane,}$ and the $xy\text{-plane}\text{.}$

Solution 解

Once again, we can almost immediately write the limits of integration and hence we can quickly proceed to evaluating the moments of inertia. Using the formula stated before, the moments of inertia of the tetrahedron $Q$ about the $xy\text{-plane,}$ the $xz\text{-plane,}$ and the $yz\text{-plane}$ are

$$\begin{array}{r} \\ {I_{x} = {\iiint\limits_{Q}{\left( {y^{2} + z^{2}} \right)\rho\left( {x,y,z} \right)dV}},} \\ {I_{y} = {\iiint\limits_{Q}{\left( {x^{2} + z^{2}} \right)\rho\left( {x,y,z} \right)dV}},} \end{array}$$

and

{I_{z} = {\iiint\limits_{Q}{\left( {x^{2} + y^{2}} \right)\rho\left( {x,y,z} \right)dV}}}\ \text{with}\ {\rho\left( {x,y,z} \right) = x^{2}yz.}$$

Proceeding with the computations, we have

$$\begin{array}{l} \\ \\ \\ \\ \\ \\ {I_{x} = {\iiint\limits_{Q}{\left( {y^{2} + z^{2}} \right)x^{2}yz\ dV}} = {\int\limits_{x = 0}^{x = 6}\ {\int\limits_{y = 0}^{y = \frac{1}{2}{({6 - x})}}\ {\int\limits_{z = 0}^{z = \frac{1}{3}{({6 - x - 2y})}}\left( {y^{2} + z^{2}} \right)}}}x^{2}yz\ dz\ dy\ dx = \frac{117}{35} \approx 3.343,} \\ {I_{y} = {\iiint\limits_{Q}{\left( {x^{2} + z^{2}} \right)x^{2}yz\ dV}} = {\int\limits_{x = 0}^{x = 6}\ {\int\limits_{y = 0}^{y = \frac{1}{2}{({6 - x})}}\ {\int\limits_{z = 0}^{z = \frac{1}{3}{({6 - x - 2y})}}\left( {x^{2} + z^{2}} \right)}}}x^{2}yz\ dz\ dy\ dx = \frac{684}{35} \approx 19.543,} \\ {I_{z} = {\iiint\limits_{Q}{\left( {x^{2} + y^{2}} \right)x^{2}yz\ dV}} = {\int\limits_{x = 0}^{x = 6}\ {\int\limits_{y = 0}^{y = \frac{1}{2}{({6 - x})}}\ {\int\limits_{z = 0}^{z = \frac{1}{3}{({6 - x - 2y})}}\left( {x^{2} + y^{2}} \right)}}}x^{2}yz\ dz\ dy\ dx = \frac{729}{35} \approx 20.829.} \end{array}$$

Thus, the moments of inertia of the tetrahedron $Q$ about the $yz\text{-plane,}$ the $xz\text{-plane,}$ and the $xy\text{-plane}$ are $117\text{/}35,684\text{/}35,\text{and}\ 729\text{/}35,$ respectively.

Consider the same region $Q$ (Figure 5.70), and use the density function $\rho\left( {x,y,z} \right) = xy^{2}z.$ Find the moments of inertia about the three coordinate planes.

Section 5.6 Exercises 5.6 节习题

In the following exercises, the region $R$ occupied by a lamina is shown in a graph. Find the mass of $R$ with the density function $\rho.$

297.

$R$ is the triangular region with vertices $\left( {0,0} \right),\left( {0,3} \right),$ and $\left( {6,0} \right);\rho\left( {x,y} \right) = xy.$

298\.

$R$ is the triangular region with vertices $\left( {0,0} \right),\left( {1,1} \right),$ $\left( {0,5} \right);\rho\left( {x,y} \right) = x + y.$

299.

$R$ is the rectangular region with vertices $\left( {0,0} \right),\left( {0,3} \right),\left( {6,3} \right),$ and $\left( {6,0} \right);$ $\rho\left( {x,y} \right) = \sqrt{xy}.$

300\.

$R$ is the rectangular region with vertices $\left( {0,1} \right),\left( {0,3} \right),\left( {3,3} \right),$ and $\left( {3,1} \right);$ $\rho\left( {x,y} \right) = x^{2}y.$

301.

$R$ is the trapezoidal region determined by the lines $y = - \frac{1}{4}x + \frac{5}{2},y = 0,y = 2,$ and $x = 0;$ $\rho\left( {x,y} \right) = 3xy.$

302\.

$R$ is the trapezoidal region determined by the lines $y = 0,y = 1,y = x,$ and $y = \text{−}x + 3;\rho\left( {x,y} \right) = 2x + y.$

303.

$R$ is the disk of radius $2$ centered at $\left( {1,2} \right);$ $\rho\left( {x,y} \right) = x^{2} + y^{2} - 2x - 4y + 5.$

304\.

$R$ is the unit disk; $\rho\left( {x,y} \right) = 3x^{4} + 6x^{2}y^{2} + 3y^{4}.$

305.

$R$ is the region enclosed by the ellipse $x^{2} + 4y^{2} = 1;\rho\left( {x,y} \right) = 1.$

306\.

$R = \left\{ {\left. \left( {x,y} \right) \right|9x^{2} + y^{2} \leq 1,x \geq 0,y \geq 0} \right\};\rho\left( {x,y} \right) = \sqrt{9x^{2} + y^{2}}.$

307.

$R$ is the region bounded by $y = x,y = \text{−}x,y = x + 2,y = \text{−}x + 2;$ $\rho\left( {x,y} \right) = 1.$

308\.

$R$ is the region bounded by $y = \frac{1}{x},y = \frac{2}{x},y = 1,$ and $y = 2;\rho\left( {x,y} \right) = 4\left( {x + y} \right).$

In the following exercises, consider a lamina occupying the region $R$ and having the density function $\rho$ given in the preceding group of exercises. Use a computer algebra system (CAS) to answer the following questions.

1. Find the moments $M_{x}$ and $M_{y}$ about the $x\text{-axis}$ and $y\text{-axis,}$ respectively.

2. Calculate and plot the center of mass of the lamina.

3. \[T\] Use a CAS to locate the center of mass on the graph of $R.$

309.

\[T\] $R$ is the triangular region with vertices $\left( {0,0} \right),\left( {0,3} \right),$ and $\left( {6,0} \right);\rho\left( {x,y} \right) = xy.$

310\.

\[T\] $R$ is the triangular region with vertices $\left( {0,0} \right),\left( {1,1} \right),\text{and}\ \left( {0,5} \right);\rho\left( {x,y} \right) = x + y.$

311.

\[T\] $R$ is the rectangular region with vertices $\left( {0,0} \right),\left( {0,3} \right),\left( {6,3} \right),\text{and}\ \left( {6,0} \right);$ $\rho\left( {x,y} \right) = \sqrt{xy}.$

312\.

\[T\] $R$ is the rectangular region with vertices $\left( {0,1} \right),\left( {0,3} \right),\left( {3,3} \right),\text{and}\ \left( {3,1} \right);$ $\rho\left( {x,y} \right) = x^{2}y.$

313.

\[T\] $R$ is the trapezoidal region determined by the lines $y = - \frac{1}{4}x + \frac{5}{2},y = 0,$ $y = 2,\text{and}\ x = 0;$ $\rho\left( {x,y} \right) = 3xy.$

314\.

\[T\] $R$ is the trapezoidal region determined by the lines $y = 0,y = 1,y = x,$ and $y = \text{−}x + 3;\rho\left( {x,y} \right) = 2x + y.$

315.

\[T\] $R$ is the disk of radius $2$ centered at $\left( {1,2} \right);$ $\rho\left( {x,y} \right) = x^{2} + y^{2} - 2x - 4y + 5.$

316\.

\[T\] $R$ is the unit disk; $\rho\left( {x,y} \right) = 3x^{4} + 6x^{2}y^{2} + 3y^{4}.$

317.

\[T\] $R$ is the region enclosed by the ellipse $x^{2} + 4y^{2} = 1;\rho\left( {x,y} \right) = 1.$

318\.

\[T\] $R = \left\{ {\left. \left( {x,y} \right) \right|9x^{2} + y^{2} \leq 1,x \geq 0,y \geq 0} \right\};\rho\left( {x,y} \right) = \sqrt{9x^{2} + y^{2}}.$

319.

\[T\] $R$ is the region bounded by $y = x,y = \text{−}x,y = x + 2,$ and $y = \text{−}x + 2;$ $\rho\left( {x,y} \right) = 1.$

320\.

\[T\] $R$ is the region bounded by $y = \frac{1}{x},$ $y = \frac{2}{x},y = 1,\text{and}\ y = 2;$ $\rho\left( {x,y} \right) = 4\left( {x + y} \right).$

In the following exercises, consider a lamina occupying the region $R$ and having the density function $\rho$ given in the first two groups of Exercises.

1. Find the moments of inertia $I_{x},I_{y},$ and $I_{0}$ about the $x\text{-axis},$ $y\text{-axis},$ and origin, respectively.

2. Find the radii of gyration with respect to the $x\text{-axis,}$ $y\text{-axis,}$ and origin, respectively.

321.

$R$ is the triangular region with vertices $\left( {0,0} \right),\left( {0,3} \right),$ and $\left( {6,0} \right);\rho\left( {x,y} \right) = xy.$

322\.

$R$ is the triangular region with vertices $\left( {0,0} \right),\left( {1,1} \right),$ and $\left( {0,5} \right);\rho\left( {x,y} \right) = x + y.$

323.

$R$ is the rectangular region with vertices $\left( {0,0} \right),\left( {0,3} \right),\left( {6,3} \right),$ and $\left( {6,0} \right);$ $\rho\left( {x,y} \right) = \sqrt{xy}.$

324\.

$R$ is the rectangular region with vertices $\left( {0,1} \right),\left( {0,3} \right),\left( {3,3} \right),$ and $\left( {3,1} \right);$ $\rho\left( {x,y} \right) = x^{2}y.$

325.

$R$ is the trapezoidal region determined by the lines $y = - \frac{1}{4}x + \frac{5}{2},y = 0,y = 2,$ and $x = 0;\rho\left( {x,y} \right) = 3xy.$

326\.

$R$ is the trapezoidal region determined by the lines $y = 0,y = 1,y = x,$ and $y = \text{−}x + 3;\rho\left( {x,y} \right) = 2x + y.$

327.

$R$ is the disk of radius $2$ centered at $\left( {1,2} \right);$ $\rho\left( {x,y} \right) = x^{2} + y^{2} - 2x - 4y + 5.$

328\.

$R$ is the unit disk; $\rho\left( {x,y} \right) = 3x^{4} + 6x^{2}y^{2} + 3y^{4}.$

329.

$R$ is the region enclosed by the ellipse $x^{2} + 4y^{2} = 1;\rho\left( {x,y} \right) = 1.$

330\.

$R = \left\{ {\left. \left( {x,y} \right) \right|9x^{2} + y^{2} \leq 1,x \geq 0,y \geq 0} \right\};\rho\left( {x,y} \right) = \sqrt{9x^{2} + y^{2}}.$

331.

$R$ is the region bounded by $y = x,y = \text{−}x,y = x + 2,\text{and}\ y = \text{−}x + 2;$ $\rho\left( {x,y} \right) = 1.$

332\.

$R$ is the region bounded by $y = \frac{1}{x},y = \frac{2}{x},y = 1,\text{and}\ y = 2;\rho\left( {x,y} \right) = 4\left( {x + y} \right).$

333.

Let $Q$ be the solid unit cube. Find the mass of the solid if its density $\rho$ is equal to the square of the distance of an arbitrary point of $Q$ to the $xy\text{-plane}.$

334\.

Let $Q$ be the solid unit hemisphere. Find the mass of the solid if its density $\rho$ is equal to the distance of an arbitrary point of $Q$ to the origin.

335.

The solid $Q$ of constant density $1$ is situated inside the sphere $x^{2} + y^{2} + z^{2} = 16$ and outside the sphere $x^{2} + y^{2} + z^{2} = 1.$ Show that the center of mass of the solid is not located within the solid.

336\.

Find the mass of the solid $Q = \left\{ {\left. \left( {x,y,z} \right) \right|1 \leq x^{2} + z^{2} \leq 25,y \leq 1 - x^{2} - z^{2}} \right\}$ whose density is $\rho\left( {x,y,z} \right) = k,$ where $k > 0.$

337.

\[T\] The solid $Q = \left\{ {\left. \left( {x,y,z} \right) \right|x^{2} + y^{2} \leq 9,0 \leq z \leq 1,x \geq 0,y \geq 0} \right\}$ has density equal to the distance to the $xy\text{-plane}\text{.}$ Use a CAS to answer the following questions.

1. Find the mass of $Q.$

2. Find the moments $M_{xy},M_{xz},\text{and}\ M_{yz}$ about the $xy\text{-plane,}$ $xz\text{-plane,}$ and $yz\text{-plane,}$ respectively.

3. Find the center of mass of $Q.$

4. Graph $Q$ and locate its center of mass.

338\.

Consider the solid $Q = \left\{ {\left. \left( {x,y,z} \right) \right|0 \leq x \leq 1,0 \leq y \leq 2,0 \leq z \leq 3} \right\}$ with the density function $\rho\left( {x,y,z} \right) = x + y + 1.$

1. Find the mass of $Q.$

2. Find the moments $M_{xy},M_{xz},\text{and}\ M_{yz}$ about the $xy\text{-plane,}$ $xz\text{-plane,}$ and $yz\text{-plane,}$ respectively.

3. Find the center of mass of $Q.$

339.

\[T\] The solid $Q$ has the mass given by the triple integral ${\int\limits_{-1}^{1}\ {\int\limits_{0}^{\frac{\pi}{4}}\ {\int\limits_{0}^{1}{r^{2}dr\ d\theta\ dz}}}}.$ Use a CAS to answer the following questions.

1. Show that the center of mass of $Q$ is located in the $xy\text{-plane.}$

2. Graph $Q$ and locate its center of mass.

340\.

The solid $Q$ is bounded by the planes $x + 4y + z = 8,x = 0,y = 0,\text{and}\ z = 0.$ Its density at any point is equal to the distance to the $xz\text{-plane}\text{.}$ Find the moment of inertia $I_{y}$ of the solid about the $xz\text{-plane}\text{.}$

341.

The solid $Q$ is bounded by the planes $x + y + z = 3,$ $x = 0,y = 0,$ and $z = 0.$ Its density is $\rho\left( {x,y,z} \right) = x + ay,$ where $a > 0.$ Show that the center of mass of the solid is located in the plane $z = \frac{3}{5}$ for any value of $a.$

342\.

Let $Q$ be the solid situated outside the sphere $x^{2} + y^{2} + z^{2} = z$ and inside the upper hemisphere $x^{2} + y^{2} + z^{2} = R^{2},$ where $R > 1.$ If the density of the solid is $\rho\left( {x,y,z} \right) = \frac{1}{\sqrt{x^{2} + y^{2} + z^{2}}},$ find $R$ such that the mass of the solid is $\frac{7\pi}{2}.$

343.

The mass of a solid $Q$ is given by ${\int\limits_{0}^{2\sqrt{2}}\ {\int\limits_{0}^{\sqrt{8 - x^{2}}}\ {\int\limits_{\sqrt{x^{2} + y^{2}}}^{\sqrt{16 - x^{2} - y^{2}}}\left( {x^{2} + y^{2} + z^{2}} \right)^{n}}}}dz\ dy\ dx,$ where $n$ is an integer. Determine $n$ such that the mass of the solid is $\left( {2 - \sqrt{2}} \right)\pi.$

344\.

Let $Q$ be the solid above the cone $x^{2} + y^{2} = z^{2}$ and below the sphere $x^{2} + y^{2} + z^{2} - 4kz = 0.$ Its density is a constant $k > 0.$ Find $k$ such that the center of mass of the solid is situated $7$ units from the origin.

345.

The solid $Q = \left\{ {\left. {(x,y,z)} \right|0 \leq x^{2} + y^{2} \leq 16,x \geq 0,y \geq 0,0 \leq z \leq x} \right\}$ has the density $\rho\left( {x,y,z} \right) = k.$ Show that the moment $M_{xy}$ about the $xy\text{-plane}$ is half of the moment $M_{yz}$ about the $yz\text{-plane}\text{.}$

346\.

The solid $Q$ is bounded by the cylinder $x^{2} + y^{2} = a^{2},$ the paraboloid $b^{2} - z = x^{2} + y^{2},$ and the $xy\text{-plane,}$ where $0 < a < b.$ Find the mass of the solid if its density is given by $\rho\left( {x,y,z} \right) = \sqrt{x^{2} + y^{2}}.$

347.

Let $Q$ be a solid of constant density $k,$ where $k > 0,$ that is located in the first octant, inside the circular cone $x^{2} + y^{2} = 9\left( {z - 1} \right)^{2},$ and above the plane $z = 0.$ Show that the moment $M_{xy}$ about the $xy\text{-plane}$ is the same as the moment $M_{yz}$ about the $yz\text{-plane}\text{.}$

348\.

The solid $Q$ has the mass given by the triple integral ${\int\limits_{0}^{1}\ {\int\limits_{0}^{\pi\text{/}2}\ {\int\limits_{0}^{r^{2}}\left( {r^{4} + r} \right)}}}dz\ d\theta\ dr.$

1. Find the density of the solid in rectangular coordinates.

2. Find the moment $M_{xy}$ about the $xy\text{-plane}\text{.}$

349.

The solid $Q$ has the moment of inertia $I_{x}$ about the $yz\text{-plane}$ given by the triple integral ${\int\limits_{0}^{2}\ {\int\limits_{\text{−}\sqrt{4 - y^{2}}}^{\sqrt{4 - y^{2}}}\ {\int\limits_{\frac{1}{2}{({x^{2} + y^{2}})}}^{\sqrt{x^{2} + y^{2}}}\left( {y^{2} + z^{2}} \right)}}}\left( {x^{2} + y^{2}} \right)dz\ dx\ dy.$

1. Find the density of $Q.$

2. Find the moment of inertia $I_{z}$ about the $xy\text{-plane.}$

350\.

The solid $Q$ has the mass given by the triple integral ${\int\limits_{0}^{\pi\text{/}4}\ {\int\limits_{0}^{2\ \text{sec}\ \theta}\ {\int\limits_{0}^{1}\left( {r^{3}\text{cos}\ \theta\ \text{sin}\ \theta + 2r} \right)}}}dz\ dr\ d\theta.$

1. Find the density of the solid in rectangular coordinates.

2. Find the moment $M_{xz}$ about the $xz\text{-plane.}$

351.

Let $Q$ be the solid bounded by the $xy\text{-plane},$ the cylinder $x^{2} + y^{2} = a^{2},$ and the plane $z = 1,$ where $a > 1$ is a real number. Find the moment $M_{xy}$ of the solid about the $xy\text{-plane}$ if its density given in cylindrical coordinates is $\rho\left( {r,\theta,z} \right) = \frac{d^{2}f}{dr^{2}}(r),$ where $f$ is a differentiable function with the first and second derivatives continuous and differentiable on $\left( {0,a} \right).$

352\.

A solid $Q$ has a volume given by $\iint\limits_{D}{\int\limits_{a}^{b}{dz\ dA,}}$ where $D$ is the projection of the solid onto the $xy\text{-plane}$ and $a < b$ are real numbers, and its density does not depend on the variable $z.$ Show that its center of mass lies in the plane $z = \frac{a + b}{2}.$

353.

Consider the solid enclosed by the cylinder $x^{2} + z^{2} = a^{2}$ and the planes $y = b$ and $y = c,$ where $a > 0$ and $b < c$ are real numbers. The density of $Q$ is given by $\rho\left( {x,y,z} \right) = f\prime(y),$ where $f$ is a differential function whose derivative is continuous on $\left( {b,c} \right).$ Show that if $f(b) = f(c),$ then the moment of inertia about the $xz\text{-plane}$ of $Q$ is null.

354\.

\[T\] The average density of a solid $Q$ is defined as $\rho_{ave} = \frac{1}{V(Q)}{\iiint\limits_{Q}{\rho\left( {x,y,z} \right)}}dV = \frac{m}{V(Q)},$ where $V(Q)$ and $m$ are the volume and the mass of $Q,$ respectively. If the density of the unit ball centered at the origin is $\rho\left( {x,y,z} \right) = e^{\text{−}x^{2} - y^{2} - z^{2}},$ use a CAS to find its average density. Round your answer to three decimal places.

355.

Show that the moments of inertia $I_{x},I_{y},\text{and}\ I_{z}$ about the $yz\text{-plane,}$ $xz\text{-plane,}$ and $xy\text{-plane,}$ respectively, of the unit ball centered at the origin whose density is $\rho\left( {x,y,z} \right) = e^{\text{−}x^{2} - y^{2} - z^{2}}$ are the same. Round your answer to two decimal places.

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5.7 Change of Variables in Multiple Integrals 5.7 多重积分的变量代换

• 5.7.1 Determine the image of a region under a given transformation of variables.
• 5.7.2 Compute the Jacobian of a given transformation.
• 5.7.3 Evaluate a double integral using a change of variables.
• 5.7.4 Evaluate a triple integral using a change of variables.

• 5.7.1 求区域在给定变量代换下的像。
• 5.7.2 计算给定变换的雅可比行列式。
• 5.7.3 用变量代换计算二重积分。
• 5.7.4 用变量代换计算三重积分。

Recall from Substitution Rule the method of integration by substitution. When evaluating an integral such as $\int_{2}^{3}{x{(x^{2} - 4)}^{5}dx,}$ we substitute $u = g(x) = x^{2} - 4.$ Then $du = 2x\ dx$ or $x\ dx = \frac{1}{2}du$ and the limits change to $u = g(2) = 2^{2} - 4 = 0$ and $u = g(3) = 9 - 4 = 5.$ Thus the integral becomes $\int_{0}^{5}{\frac{1}{2}u^{5}du}$ and this integral is much simpler to evaluate. In other words, when solving integration problems, we make appropriate substitutions to obtain an integral that becomes much simpler than the original integral.

We also used this idea when we transformed double integrals in rectangular coordinates to polar coordinates and transformed triple integrals in rectangular coordinates to cylindrical or spherical coordinates to make the computations simpler. More generally,

$${\int\limits_{a}^{b}{f(x)dx =}}{\int\limits_{c}^{d}{f\left( {g(u)} \right)}}g\prime(u)du,$$

Where $x = g(u),dx = g\prime(u)du,$ and $u = c$ and $u = d$ satisfy $c = g^{- 1}(a)$ and $d = g^{- 1}(b).$

A similar result occurs in double integrals when we substitute $x = h\left( {r,\theta} \right) = r\ \text{cos}\ \theta,$ $y = g\left( {r,\theta} \right) = r\ \text{sin}\ \theta,$ and $dA = dx\ dy = r\ dr\ d\theta.$ Then we get

$${\iint\limits_{R}{f\left( {x,y} \right)}}dA = {\iint\limits_{S}{f\left( {r\ \text{cos}\ \theta,r\ \text{sin}\ \theta} \right)r\ dr\ d\theta}}$$

where the domain $R$ is replaced by the domain $S$ in polar coordinates. Generally, the function that we use to change the variables to make the integration simpler is called a transformation or mapping.

Planar Transformations 平面变换

A planar transformation $T$ is a function that transforms a region $G$ in one plane into a region $R$ in another plane by a change of variables. Both $G$ and $R$ are subsets of $\mathbb{R}^{2}.$ For example, Figure 5.71 shows a region $G$ in the $uv\text{-plane}$ transformed into a region $R$ in the $xy\text{-plane}$ by the change of variables $x = g\left( {u,v} \right)$ and $y = h\left( {u,v} \right),$ or sometimes we write $x = x\left( {u,v} \right)$ and $y = y\left( {u,v} \right).$ We shall typically assume that each of these functions has continuous first partial derivatives, which means $g_{u},g_{v},h_{u},$ and $h_{v}$ exist and are also continuous. The need for this requirement will become clear soon.

A transformation $T\text{:}\ G\rightarrow R,$ defined as $T\left( {u,v} \right) = \left( {x,y} \right),$ is said to be a one-to-one transformation if no two points map to the same image point.

To show that $T$ is a one-to-one transformation, we assume $T\left( {u_{1},v_{1}} \right) = T\left( {u_{2},v_{2}} \right)$ and show that as a consequence we obtain $\left( {u_{1},v_{1}} \right) = \left( {u_{2},v_{2}} \right).$ If the transformation $T$ is one-to-one in the domain $G,$ then the inverse $T^{-1}$ exists with the domain $R$ such that $T^{-1} \circ T$ and $T \circ T^{-1}$ are identity functions.

Figure 5.71 shows the mapping $T\left( {u,v} \right) = \left( {x,y} \right)$ where $x$ and $y$ are related to $u$ and $v$ by the equations $x = g\left( {u,v} \right)$ and $y = h\left( {u,v} \right).$ The region $G$ is the domain of $T$ and the region $R$ is the range of $T,$ also known as the *image* of $G$ under the transformation $T.$

Determining How the Transformation Works 弄清变换的作用方式

Suppose a transformation $T$ is defined as $T\left( {r,\theta} \right) = \left( {x,y} \right)$ where $x = r\ \text{cos}\ \theta,y = r\ \text{sin}\ \theta.$ Find the image of the polar rectangle $G = \left\{ {\left. \left( {r,\theta} \right) \right|0 < r \leq 1,0 \leq \theta \leq \pi\text{/}2} \right\}$ in the $r\theta\text{-plane}$ to a region $R$ in the $xy\text{-plane}\text{.}$ Show that $T$ is a one-to-one transformation in $G$ and find $T^{-1}\left( {x,y} \right).$

Solution 解

Since $r$ varies from 0 to 1 in the $r\theta\text{-plane},$ we have a circular disc of radius 0 to 1 in the $xy\text{-plane}\text{.}$ Because $\theta$ varies from 0 to $\pi\text{/2}$ in the $r\theta\text{-plane},$ we end up getting a quarter circle of radius $1$ in the first quadrant of the $xy\text{-plane}$ (Figure 5.72). Hence $R$ is a quarter circle bounded by $x^{2} + y^{2} = 1$ in the first quadrant.

In order to show that $T$ is a one-to-one transformation, assume $T\left( {r_{1},\theta_{1}} \right) = T\left( {r_{2},\theta_{2}} \right)$ and show as a consequence that $\left( {r_{1},\theta_{1}} \right) = \left( {r_{2},\theta_{2}} \right).$ In this case, we have

$$\begin{matrix} {T\left( r_{1},\theta_{1} \right)} & = & {T\left( r_{2},\theta_{2} \right),} \\ \left( x_{1},y_{1} \right) & = & {\left( x_{2},y_{2} \right),} \\ \left( r_{1}\text{cos}\ \theta_{1},r_{1}\text{sin}\ \theta_{1} \right) & = & {\left( r_{2}\text{cos}\ \theta_{2},r_{2}\text{sin}\ \theta_{2} \right),} \\ {r_{1}\text{cos}\ \theta_{1}} & = & {r_{2}\text{cos}\ \theta_{2}} \\ {r_{1}\text{sin}\ \theta_{1}} & = & {r_{2}\text{sin}\ \theta_{2}} \end{matrix}$$

Dividing, we obtain

$$\begin{array}{rll} \frac{r_{1}\text{cos}\ \theta_{1}}{r_{1}\text{sin}\ \theta_{1}} & = & \frac{r_{2}\text{cos}\ \theta_{2}}{r_{2}\text{sin}\ \theta_{2}} \\ \frac{\text{cos}\ \theta_{1}}{\text{sin}\ \theta_{1}} & = & \frac{\text{cos}\ \theta_{2}}{\text{sin}\ \theta_{2}} \\ {\text{cot}\ \theta_{1}} & = & {\text{cot}\ \theta_{2}} \\ \theta_{1} & = & \theta_{2} \end{array}$$

since the cotangent function is one-one function in the interval $0 \leq \theta \leq \pi\text{/}2.$ Also, since $0 < r \leq 1,$ we have $r_{1} = r_{2},\theta_{1} = \theta_{2}.$ Therefore, $\left( {r_{1},\theta_{1}} \right) = \left( {r_{2},\theta_{2}} \right)$ and $T$ is a one-to-one transformation from $G$ into $R.$

To find $T^{-1}\left( {x,y} \right)$ solve for $r,\theta$ in terms of $x,y.$ We already know that $r^{2} = x^{2} + y^{2}$ and $\text{tan}\ \theta = \frac{y}{x}.$ Thus $T^{-1}\left( {x,y} \right) = \left( {r,\theta} \right)$ is defined as $r = \sqrt{x^{2} + y^{2}}$ and $\theta = \text{tan}^{-1}\left( \frac{y}{x} \right).$

Finding the Image under $T$ 求 $T$ 下的像

Let the transformation $T$ be defined by $T\left( {u,v} \right) = \left( {x,y} \right)$ where $x = u^{2} - v^{2}$ and $y = uv.$ Find the image of the triangle in the $uv\text{-plane}$ with vertices $\left( {0,0} \right),\left( {0,1} \right),$ and $\left( {1,1} \right).$

Solution 解

The triangle and its image are shown in Figure 5.73. To understand how the sides of the triangle transform, call the side that joins $\left( {0,0} \right)$ and $\left( {0,1} \right)$ side $A,$ the side that joins $\left( {0,0} \right)$ and $\left( {1,1} \right)$ side $B,$ and the side that joins $\left( {1,1} \right)$ and $\left( {0,1} \right)$ side $C.$

For the side $A\text{:}\ u = 0,0 \leq v \leq 1$ transforms to $x = \text{−}v^{2},y = 0$ so this is the side $A\prime$ that joins $\left( {-1,0} \right)$ and $\left( {0,0} \right).$

For the side $B\text{:}\ u = v,0 \leq u \leq 1$ transforms to $x = 0,y = u^{2}$ so this is the side $B^{\prime}$ that joins $\left( {0,0} \right)$ and $\left( {0,1} \right).$

For the side $C\text{:}\ 0 \leq u \leq 1,v = 1$ transforms to $x = u^{2} - 1,y = u$ (hence $x = y^{2} - 1)$ so this is the side $C^{\prime}$ that makes the upper half of the parabolic arc joining $\left( {-1,0} \right)$ and $\left( {0,1} \right).$

All the points in the entire region of the triangle in the $uv\text{-plane}$ are mapped inside the parabolic region in the $xy\text{-plane}\text{.}$

Let a transformation $T$ be defined as $T\left( {u,v} \right) = \left( {x,y} \right)$ where $x = u + v,y = 3v.$ Find the image of the rectangle $G = \left\{ {\left( {u,v} \right)\text{:}\ 0 \leq u \leq 1,0 \leq v \leq 2} \right\}$ from the $uv\text{-plane}$ after the transformation into a region $R$ in the $xy\text{-plane}\text{.}$ Show that $T$ is a one-to-one transformation and find $T^{-1}\left( {x,y} \right).$

Jacobians 雅可比行列式

Recall that we mentioned near the beginning of this section that each of the component functions must have continuous first partial derivatives, which means that $g_{u},g_{v},h_{u},$ and $h_{v}$ exist and are also continuous. A transformation that has this property is called a $C^{1}$ transformation (here $C$ denotes continuous). Let $T\left( {u,v} \right) = \left( {g\left( {u,v} \right),h\left( {u,v} \right)} \right),$ where $x = g\left( {u,v} \right)$ and $y = h\left( {u,v} \right),$ be a one-to-one $C^{1}$ transformation. We want to see how it transforms a small rectangular region $S,$ $\text{Δ}u$ units by $\text{Δ}v$ units, in the $uv\text{-plane}$ (see the following figure).

Since $x = g\left( {u,v} \right)$ and $y = h\left( {u,v} \right),$ we have the position vector $\mathbf{r}\left( {u,v} \right) = g\left( {u,v} \right)\mathbf{i} + h\left( {u,v} \right)\mathbf{j}$ of the image of the point $\left( {u,v} \right).$ Suppose that $\left( {u_{0},v_{0}} \right)$ is the coordinate of the point at the lower left corner that mapped to $\left( {x_{0},y_{0}} \right) = T\left( {u_{0},v_{0}} \right).$ The line $v = v_{0}$ maps to the image curve with vector function $\text{r}\left( {u,v_{0}} \right),$ and the tangent vector at $\left( {x_{0},y_{0}} \right)$ to the image curve is

$$\mathbf{r}_{u} = g_{u}\left( {u_{0},v_{0}} \right)\mathbf{i} + h_{u}\left( {u_{0},v_{0}} \right)\mathbf{j} = \frac{\partial x}{\partial u}\mathbf{i} + \frac{\partial y}{\partial u}\mathbf{j}.$$

Similarly, the line $u = u_{0}$ maps to the image curve with vector function $\mathbf{r}\left( {u_{0},v} \right),$ and the tangent vector at $\left( {x_{0},y_{0}} \right)$ to the image curve is

$$\mathbf{r}_{v} = g_{v}\left( {u_{0},v_{0}} \right)\mathbf{i} + h_{v}\left( {u_{0},v_{0}} \right)\mathbf{j} = \frac{\partial x}{\partial v}\mathbf{i} + \frac{\partial y}{\partial v}\mathbf{j}.$$

Now, note that

$$\mathbf{r}_{u} = \underset{\text{Δ}u\rightarrow 0}{\text{lim}}\frac{\mathbf{r}\left( {u_{0} + \text{Δ}u,v_{0}} \right) - \mathbf{r}\left( {u_{0},v_{0}} \right)}{\text{Δ}u}\ \text{so}\ \mathbf{r}\left( {u_{0} + \text{Δ}u,v_{0}} \right) - \mathbf{r}\left( {u_{0},v_{0}} \right) \approx \text{Δ}u\mathbf{r}_{u}.$$

Similarly,

$$\mathbf{r}_{v} = \underset{\text{Δ}v\rightarrow 0}{\text{lim}}\frac{\mathbf{r}\left( {u_{0},v_{0} + \text{Δ}v} \right) - \mathbf{r}\left( {u_{0},v_{0}} \right)}{\text{Δ}v}\ \text{so}\ \mathbf{r}\left( {u_{0},v_{0} + \text{Δ}v} \right) - \mathbf{r}\left( {u_{0},v_{0}} \right) \approx \text{Δ}v\mathbf{r}_{v}.$$

This allows us to estimate the area $\text{Δ}A$ of the image $R$ by finding the area of the parallelogram formed by the sides $\text{Δ}v\mathbf{r}_{v}$ and $\text{Δ}u\mathbf{r}_{u}.$ By using the cross product of these two vectors by adding the k component as $0,$ the area $\text{Δ}A$ of the image $R$ (refer to The Cross Product) is approximately $\mathbf{\left. ||{\text{Δ}{ur}_{u}\ \times \ \text{Δ}{vr}_{v}} \right.||} = \mathbf{\left. ||{r_{u}\ \times \ r_{v}} \right.||}\text{Δ}u\text{Δ}v.$ In determinant form, the cross product is

$$\mathbf{r}_{u}\ \times \ \mathbf{r}_{v} = \left| \begin{matrix} \mathbf{i} & & \mathbf{j} & & \mathbf{k} \\ \frac{\partial x}{\partial u} & & \frac{\partial y}{\partial u} & & 0 \\ \frac{\partial x}{\partial v} & & \frac{\partial y}{\partial v} & & 0 \end{matrix} \right| = \left| \begin{array}{lll} \frac{\partial x}{\partial u} & & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & & \frac{\partial y}{\partial v} \end{array} \right|\mathbf{k} = \left( {\frac{\partial x}{\partial u}\ \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\ \frac{\partial y}{\partial u}} \right)\mathbf{k}.$$

Since $\mathbf{\left. ||k \right.||} = 1,$ we have $\text{Δ}A \approx \mathbf{\left. ||{r_{u}\ \times \ r_{v}} \right.||}\text{Δ}u\text{Δ}v = \left( \frac{\partial x}{\partial u}\ \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\ \frac{\partial y}{\partial u} \right)\text{Δ}u\text{Δ}v.$

The Jacobian of the $C^{1}$ transformation $T\left( {u,v} \right) = \left( {g\left( {u,v} \right),h\left( {u,v} \right)} \right)$ is denoted by $J\left( {u,v} \right)$ and is defined by the $2\ \times \ 2$ determinant

$$J\left( {u,v} \right) = \left| \frac{\partial\left( {x,y} \right)}{\partial\left( {u,v} \right)} \right| = \left| \begin{array}{lll} \frac{\partial x}{\partial u} & & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & & \frac{\partial y}{\partial v} \end{array} \right| = \left( {\frac{\partial x}{\partial u}\ \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\ \frac{\partial y}{\partial u}} \right).$$

Using the definition, we have

$$\text{Δ}A \approx J\left( {u,v} \right)\text{Δ}u\text{Δ}v = \left| \frac{\partial\left( {x,y} \right)}{\partial\left( {u,v} \right)} \right|\text{Δ}u\text{Δ}v.$$

Note that the Jacobian is frequently denoted simply by

$$J\left( {u,v} \right) = \frac{\partial\left( {x,y} \right)}{\partial\left( {u,v} \right)}.$$

Note also that

$$\left| \begin{array}{lll} \frac{\partial x}{\partial u} & & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & & \frac{\partial y}{\partial v} \end{array} \right| = \left( {\frac{\partial x}{\partial u}\ \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\ \frac{\partial y}{\partial u}} \right) = \left| \begin{array}{lll} \frac{\partial x}{\partial u} & & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & & \frac{\partial y}{\partial v} \end{array} \right|.$$

Hence the notation $J\left( {u,v} \right) = \frac{\partial\left( {x,y} \right)}{\partial\left( {u,v} \right)}$ suggests that we can write the Jacobian determinant with partials of $x$ in the first row and partials of $y$ in the second row.

Finding the Jacobian 求雅可比行列式

Find the Jacobian of the transformation given in Example 5.65.

Solution 解

The transformation in the example is $T\left( {r,\theta} \right) = \left( {r\ \text{cos}\ \theta,r\ \text{sin}\ \theta} \right)$ where $x = r\ \text{cos}\ \theta$ and $y = r\ \text{sin}\ \theta.$ Thus the Jacobian is

$$\begin{array}{cl} {J\left( {r,\theta} \right)} & {= \frac{\partial\left( {x,y} \right)}{\partial\left( {r,\theta} \right)} = \left| \begin{array}{lll} \frac{\partial x}{\partial r} & & \frac{\partial x}{\partial\theta} \\ \frac{\partial y}{\partial r} & & \frac{\partial y}{\partial\theta} \end{array} \right| = \left| \begin{array}{llr} {\text{cos}\ \theta} & & {- r\ \text{sin}\ \theta} \\ {\text{sin}\ \theta} & & {r\ \text{cos}\ \theta} \end{array} \right|} \\ & {= r\ \text{cos}^{2}\theta + r\ \text{sin}^{2}\theta = r\left( {\text{cos}^{2}\theta + \text{sin}^{2}\theta} \right) = r.} \end{array}$$

Finding the Jacobian 求雅可比行列式

Find the Jacobian of the transformation given in Example 5.66.

Solution 解

The transformation in the example is $T\left( {u,v} \right) = \left( {u^{2} - v^{2},uv} \right)$ where $x = u^{2} - v^{2}$ and $y = uv.$ Thus the Jacobian is

$$J\left( {u,v} \right) = \frac{\partial\left( {x,y} \right)}{\partial\left( {u,v} \right)} = \left| \begin{array}{lll} \frac{\partial x}{\partial u} & & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & & \frac{\partial y}{\partial v} \end{array} \right| = \left| \begin{array}{lll} {2u} & & v \\ {- 2v} & & u \end{array} \right| = 2u^{2} + 2v^{2}.$$

Find the Jacobian of the transformation: $T\left( {u,v} \right) = \left( {u + v,2v} \right).$

Change of Variables for Double Integrals 二重积分的变量代换

We have already seen that, under the change of variables $T\left( {u,v} \right) = \left( {x,y} \right)$ where $x = g\left( {u,v} \right)$ and $y = h\left( {u,v} \right),$ a small region $\text{Δ}A$ in the $xy\text{-plane}$ is related to the area formed by the product $\text{Δ}u\text{Δ}v$ in the $uv\text{-plane}$ by the approximation

$$\text{Δ}A \approx J\left( {u,v} \right)\text{Δ}u,\text{Δ}v.$$

Now let’s go back to the definition of double integral for a minute:

$${\iint\limits_{R}{f\left( {x,y} \right)dA = \underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f\left( {x_{ij},y_{ij}} \right)}}}}}\text{Δ}A.$$

Referring to Figure 5.75, observe that we divided the region $S$ in the $uv\text{-plane}$ into small subrectangles $S_{ij}$ and we let the subrectangles $R_{ij}$ in the $xy\text{-plane}$ be the images of $S_{ij}$ under the transformation $T\left( {u,v} \right) = \left( {x,y} \right).$

Then the double integral becomes

$${\iint\limits_{R}{f\left( {x,y} \right)dA = \underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f\left( {x_{ij},y_{ij}} \right)}}}}}\text{Δ}A = \underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f\left( {g\left( {u_{ij},v_{ij}} \right),h\left( {u_{ij},v_{ij}} \right)} \right)}}}\left| {J\left( {u_{ij},v_{ij}} \right)} \right|\text{Δ}u\text{Δ}v.$$

Notice this is exactly the double Riemann sum for the integral

$${\iint\limits_{S}{f\left( {g\left( {u,v} \right),h\left( {u,v} \right)} \right)}}\left| \frac{\partial\left( {x,y} \right)}{\partial\left( {u,v} \right)} \right|du\ dv.$$

Change of Variables for Double Integrals 二重积分的变量代换

Let $T\left( {u,v} \right) = \left( {x,y} \right)$ where $x = g\left( {u,v} \right)$ and $y = h\left( {u,v} \right)$ be a one-to-one $C^{1}$ transformation, with a nonzero Jacobian on the interior of the region $S$ in the $uv\text{-plane;}$ it maps $S$ into the region $R$ in the $xy\text{-plane}\text{.}$ If $f$ is continuous on $R,$ then

$${\iint\limits_{R}{f\left( {x,y} \right)}}dA = {\iint\limits_{S}{f\left( {g\left( {u,v} \right),h\left( {u,v} \right)} \right)}}\left| \frac{\partial\left( {x,y} \right)}{\partial\left( {u,v} \right)} \right|du\ dv.$$

With this theorem for double integrals, we can change the variables from $\left( {x,y} \right)$ to $\left( {u,v} \right)$ in a double integral simply by replacing

$$dA = dx\ dy = \left| \frac{\partial\left( {x,y} \right)}{\partial\left( {u,v} \right)} \right|du\ dv$$

when we use the substitutions $x = g\left( {u,v} \right)$ and $y = h\left( {u,v} \right)$ and then change the limits of integration accordingly. This change of variables often makes any computations much simpler.

Changing Variables from Rectangular to Polar Coordinates 由直角坐标换为极坐标

Consider the integral

$${\int\limits_{0}^{2}\ {\int\limits_{0}^{\sqrt{2x - x^{2}}}{\sqrt{x^{2} + y^{2}}\ dy\ dx}}}.$$

Use the change of variables $x = r\ \text{cos}\ \theta$ and $y = r\ \text{sin}\ \theta,$ and find the resulting integral.

Solution 解

First we need to find the region of integration. This region is bounded below by $y = 0$ and above by $y = \sqrt{2x - x^{2}}$ (see the following figure).

Squaring and collecting terms, we find that the region is the upper half of the circle $x^{2} + y^{2} - 2x = 0,$ that is, $y^{2} + \left( {x - 1} \right)^{2} = 1.$ In polar coordinates, the circle is $r = 2\ \text{cos}\ \theta$ so the region of integration in polar coordinates is bounded by $0 \leq r \leq 2~\text{cos}\ \theta$ and $0 \leq \theta \leq \frac{\pi}{2}.$

The Jacobian is $J\left( {r,\theta} \right) = r,$ as shown in Example 5.67. Since $r \geq 0,$ we have $\left| {J\left( {r,\theta} \right)} \right| = r.$

The integrand $\sqrt{x^{2} + y^{2}}$ changes to $r$ in polar coordinates, so the double iterated integral is

$${\int\limits_{0}^{2}\ {\int\limits_{0}^{\sqrt{2x - x^{2}}}\sqrt{x^{2} + y^{2}}}}dy\ dx = {\int\limits_{0}^{\pi\text{/}2}\ {\int\limits_{0}^{2\ \text{cos}\ \theta}{r\left| {J\left( {r,\theta} \right)} \right|}}}dr\ d\theta = {\int\limits_{0}^{\pi\text{/}2}\ {\int\limits_{0}^{2\ \text{cos}\ \theta}r^{2}}}dr\ d\theta.$$

Considering the integral ${\int\limits_{0}^{1}\ {\int\limits_{0}^{\sqrt{1 - x^{2}}}\left( {x^{2} + y^{2}} \right)}}dy\ dx,$ use the change of variables $x = r\ \text{cos}\ \theta$ and $y = r\ \text{sin}\ \theta,$ and find the resulting integral.

Notice in the next example that the region over which we are to integrate may suggest a suitable transformation for the integration. This is a common and important situation.

Changing Variables 变量代换

Consider the integral ${\iint\limits_{R}\left( {x - y} \right)}dy\ dx,$ where $R$ is the parallelogram joining the points $\left( {1,2} \right),$ $\left( {3,4} \right),\left( {4,3} \right),$ and $\left( {6,5} \right)$ (Figure 5.77). Make appropriate changes of variables, and write the resulting integral.

Solution 解

First, we need to understand the region over which we are to integrate. The sides of the parallelogram are $x - y + 1 = 0,x - y - 1 = 0,$ $x - 3y + 5 = 0,\text{and}\ x - 3y + 9 = 0$ (Figure 5.78). Another way to look at them is $x - y = -1,x - y = 1,$ $x - 3y = -5,$ and $x - 3y = - 9.$

Clearly the parallelogram is bounded by the lines $y = x + 1,y = x - 1,y = \frac{1}{3}\left( {x + 5} \right),$ and $y = \frac{1}{3}\left( {x + 9} \right).$

Notice that if we were to make $u = x - y$ and $v = x - 3y,$ then the limits on the integral would be $-1 \leq u \leq 1$ and $-9 \leq v \leq - 5.$

To solve for $x$ and $y,$ we multiply the first equation by $3$ and subtract the second equation, $3u - v = \left( {3x - 3y} \right) - \left( {x - 3y} \right) = 2x.$ Then we have $x = \frac{3u - v}{2}.$ Moreover, if we simply subtract the second equation from the first, we get $u - v = \left( {x - y} \right) - \left( {x - 3y} \right) = 2y$ and $y = \frac{u - v}{2}.$

Thus, we can choose the transformation

$$T\left( {u,v} \right) = \left( {\frac{3u - v}{2},\frac{u - v}{2}} \right)$$

and compute the Jacobian $J\left( {u,v} \right).$ We have

$$J\left( {u,v} \right) = \frac{\partial\left( {x,y} \right)}{\partial\left( {u,v} \right)} = \left| \begin{array}{lll} \frac{\partial x}{\partial u} & & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & & \frac{\partial y}{\partial v} \end{array} \right| = \left| \begin{array}{lll} {3\text{/}2} & & {- 1\text{/}2} \\ {1\text{/}2} & & {- 1\text{/}2} \end{array} \right| = - \frac{3}{4} + \frac{1}{4} = - \frac{1}{2}.$$

Therefore, $\left| {J\left( {u,v} \right)} \right| = \frac{1}{2}.$ Also, the original integrand becomes

$$x - y = \frac{1}{2}\left\lbrack {3u - v - u + v} \right\rbrack = \frac{1}{2}\left\lbrack {3u - u} \right\rbrack = \frac{1}{2}\left\lbrack {2u} \right\rbrack = u.$$

Therefore, by the use of the transformation $T,$ the integral changes to

$${\iint\limits_{R}{\left( {x - y} \right)dy\ dx = {\int\limits_{-9}^{-5}\ {\int\limits_{-1}^{1}{J\left( {u,v} \right)u\ du\ dv = {\int\limits_{-9}^{-5}\ {\int\limits_{-1}^{1}\left( \frac{1}{2} \right)}}}}}}}u\ du\ dv,$$

which is much simpler to compute. In fact, it is easily found to be zero. And this is just one example of why we transform integrals like this.

Make appropriate changes of variables in the integral ${\iint\limits_{R}\frac{4}{\left( {x - y} \right)^{2}}}dy\ dx,$ where $R$ is the trapezoid bounded by the lines $x - y = 2,x - y = 4,x = 0,\text{and}\ y = 0.$ Write the resulting integral.

We are ready to give a problem-solving strategy for change of variables.

Change of Variables 变量代换

1. Sketch the region given by the problem in the $xy\text{-plane}$ and then write the equations of the curves that form the boundary.

2. Depending on the region or the integrand, choose the transformations $x = g\left( {u,v} \right)$ and $y = h\left( {u,v} \right).$

3. Determine the new limits of integration in the $uv\text{-plane}\text{.}$

4. Find the Jacobian $J\left( {u,v} \right).$

5. In the integrand, replace the variables to obtain the new integrand.

6. Replace $dy\ dx$ or $dx\ dy,$ whichever occurs, by $J\left( {u,v} \right)du\ dv.$

In the next example, we find a substitution that makes the integrand much simpler to compute.

Evaluating an Integral 计算积分

Using the change of variables $u = x - y$ and $v = x + y,$ evaluate the integral

$${\iint\limits_{R}{\left( {x - y} \right)e^{x^{2} - y^{2}}dA}},$$

where $R$ is the region bounded by the lines $x + y = 1$ and $x + y = 3$ and the curves $x^{2} - y^{2} = -1$ and $x^{2} - y^{2} = 1$ (see the first region in Figure 5.79).

Solution 解

As before, first find the region $R$ and picture the transformation so it becomes easier to obtain the limits of integration after the transformations are made (Figure 5.79).

Given $u = x - y$ and $v = x + y,$ we have $x = \frac{u + v}{2}$ and $y = \frac{v - u}{2}$ and hence the transformation to use is $T\left( {u,v} \right) = \left( {\frac{u + v}{2},\frac{v - u}{2}} \right).$ The lines $x + y = 1$ and $x + y = 3$ become $v = 1$ and $v = 3,$ respectively. The curves $x^{2} - y^{2} = 1$ and $x^{2} - y^{2} = -1$ become $uv = 1$ and $uv = -1,$ respectively.

Thus we can describe the region $S$ (see the second region Figure 5.79) as

$$S = \left\{ {\left. \left( {u,v} \right) \right|1 \leq v \leq 3,\frac{-1}{v} \leq u \leq \frac{1}{v}} \right\}.$$

The Jacobian for this transformation is

$$J\left( {u,v} \right) = \frac{\partial\left( {x,y} \right)}{\partial\left( {u,v} \right)} = \left| \begin{array}{lll} \frac{\partial x}{\partial u} & & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & & \frac{\partial y}{\partial v} \end{array} \right| = \left| \begin{array}{lll} {1\text{/}2} & & {- 1\text{/}2} \\ {1\text{/}2} & & {1\text{/}2} \end{array} \right| = \frac{1}{2}.$$

Therefore, by using the transformation $T,$ the integral changes to

$${\iint\limits_{R}\left( {x - y} \right)}e^{x^{2} - y^{2}}dA = \frac{1}{2}{\int\limits_{1}^{3}\ {\int\limits_{-1\text{/}v}^{1\text{/}v}{ue^{uv}du\ dv}}}.$$

Doing the evaluation, we have

$$\frac{1}{2}{\int\limits_{1}^{3}\ {\int\limits_{-1\text{/}v}^{1\text{/}v}{ue^{uv}du\ dv}}} = \frac{2}{3e} \approx 0.245.$$

Using the substitutions $x = v$ and $y = \sqrt{u + v},$ evaluate the integral ${\iint\limits_{R}{y\ \text{sin}\left( {y^{2} - x} \right)}}dA$ where $R$ is the region bounded by the lines $y = \sqrt{x},x = 2,\text{and}\ y = 0.$

Change of Variables for Triple Integrals 三重积分的变量代换

Changing variables in triple integrals works in exactly the same way. Cylindrical and spherical coordinate substitutions are special cases of this method, which we demonstrate here.

Suppose that $G$ is a region in $uvw\text{-space}$ and is mapped to $D$ in $xyz\text{-space}$ (Figure 5.80) by a one-to-one $C^{1}$ transformation $T\left( {u,v,w} \right) = \left( {x,y,z} \right)$ where $x = g\left( {u,v,w} \right),$ $y = h\left( {u,v,w} \right),$ and $z = k\left( {u,v,w} \right).$

Then any function $F\left( {x,y,z} \right)$ defined on $D$ can be thought of as another function $H\left( {u,v,w} \right)$ that is defined on $G\text{:}$

$$F\left( {x,y,z} \right) = F\left( {g\left( {u,v,w} \right),h\left( {u,v,w} \right),k\left( {u,v,w} \right)} \right) = H\left( {u,v,w} \right).$$

Now we need to define the Jacobian for three variables.

The Jacobian determinant $J\left( {u,v,w} \right)$ in three variables is defined as follows:

$$J(u,v,w) = \left| \begin{array}{lclcl} \frac{\partial x}{\partial u} & & \frac{\partial y}{\partial u} & & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & & \frac{\partial y}{\partial v} & & \frac{\partial z}{\partial v} \\ \frac{\partial x}{\partial w} & & \frac{\partial y}{\partial w} & & \frac{\partial z}{\partial w} \end{array} \right|.$$

This is also the same as

$$J(u,v,w) = \left| \begin{array}{lclcl} \frac{\partial x}{\partial u} & & \frac{\partial x}{\partial v} & & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & & \frac{\partial y}{\partial v} & & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & & \frac{\partial z}{\partial v} & & \frac{\partial z}{\partial w} \end{array} \right|.$$

The Jacobian can also be simply denoted as $\frac{\partial\left( {x,y,z} \right)}{\partial\left( {u,v,w} \right)}.$

With the transformations and the Jacobian for three variables, we are ready to establish the theorem that describes change of variables for triple integrals.

Change of Variables for Triple Integrals 三重积分的变量代换

Let $T\left( {u,v,w} \right) = \left( {x,y,z} \right)$ where $x = g\left( {u,v,w} \right),y = h\left( {u,v,w} \right),$ and $z = k\left( {u,v,w} \right),$ be a one-to-one $C^{1}$ transformation, with a nonzero Jacobian, that maps the region $G$ in the $uvw\text{-space}$ into the region $R$ in the $xyz\text{-space}\text{.}$ As in the two-dimensional case, if $F$ is continuous on $R,$ then

$$\begin{array}{cl} {\iiint\limits_{R}{F\left( {x,y,z} \right)dV}} & {= {\iiint\limits_{G}{F\left( {g\left( {u,v,w} \right),h\left( {u,v,w} \right),k\left( {u,v,w} \right)} \right)\left| \frac{\partial\left( {x,y,z} \right)}{\partial\left( {u,v,w} \right)} \right|}}du\ dv\ dw} \\ & {= {\iiint\limits_{G}{H\left( {u,v,w} \right)}}\left| {J\left( {u,v,w} \right)} \right|du\ dv\ dw.} \end{array}$$

Let us now see how changes in triple integrals for cylindrical and spherical coordinates are affected by this theorem. We expect to obtain the same formulas as in Triple Integrals in Cylindrical and Spherical Coordinates.

Obtaining Formulas in Triple Integrals for Cylindrical and Spherical Coordinates 三重积分中柱坐标与球坐标公式的推导

Derive the formula in triple integrals for

1. cylindrical and

2. spherical coordinates.

Solution 解

1. For cylindrical coordinates, the transformation is $T\left( {r,\theta,z} \right) = \left( {x,y,z} \right)$ from the Cartesian $r\theta z\text{-plane}$ to the Cartesian $xyz\text{-plane}$ (Figure 5.81). Here $x = r\ \text{cos}\ \theta,$ $y = r\ \text{sin}\ \theta,$ and $z = z.$ The Jacobian for the transformation is

$$\begin{array}{cl} {J\left( {r,\theta,z} \right)} & {= \frac{\partial\left( {x,y,z} \right)}{\partial\left( {r,\theta,z} \right)} = \left| \begin{array}{lllll} \frac{\partial x}{\partial r} & & \frac{\partial x}{\partial\theta} & & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial r} & & \frac{\partial y}{\partial\theta} & & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial r} & & \frac{\partial z}{\partial\theta} & & \frac{\partial z}{\partial z} \end{array} \right|} \\ & {= \left| \begin{array}{clcll} {\text{cos}\ \theta} & & {- r\ \text{sin}\ \theta} & & 0 \\ {\text{sin}\ \theta} & & {r\ \text{cos}\ \theta} & & 0 \\ 0 & & 0 & & 1 \end{array} \right| = r\ \text{cos}^{2}\theta + r\ \text{sin}^{2}\theta = r\left( {\text{cos}^{2}\theta + \text{sin}^{2}\theta} \right) = r.} \end{array}$$

We know that $r \geq 0,$ so $\left| {J\left( {r,\theta,z} \right)} \right| = r.$ Then the triple integral is

$${\iiint\limits_{D}{f\left( {x,y,z} \right)dV = {\iiint\limits_{G}{f\left( {r\ \text{cos}\ \theta,r\ \text{sin}\ \theta,z} \right)}}}}r\ dr\ d\theta\ dz.$$

2. For spherical coordinates, the transformation is $T\left( {\rho,\theta,\varphi} \right) = \left( {x,y,z} \right)$ from the Cartesian $p\theta\varphi\text{-plane}$ to the Cartesian $xyz\text{-plane}$ (Figure 5.82). Here $x = \rho\ \text{sin}\ \varphi\ \text{cos}\ \theta,$ $y = \rho\ \text{sin}\ \varphi\ \text{sin}\ \theta,$ and $z = \rho\ \text{cos}\ \varphi.$ The Jacobian for the transformation is

$$J\left( {\rho,\theta,\varphi} \right) = \frac{\partial\left( {x,y,z} \right)}{\partial\left( {\rho,\theta,\varphi} \right)} = \left| \begin{array}{lllll} \frac{\partial x}{\partial\rho} & & \frac{\partial x}{\partial\theta} & & \frac{\partial x}{\partial\varphi} \\ \frac{\partial y}{\partial\rho} & & \frac{\partial y}{\partial\theta} & & \frac{\partial y}{\partial\varphi} \\ \frac{\partial z}{\partial\rho} & & \frac{\partial z}{\partial\theta} & & \frac{\partial z}{\partial\varphi} \end{array} \right| = \left| \begin{array}{clcll} {\text{sin}\ \varphi\ \text{cos}\ \theta} & & {\rho\ \text{sin}\ \varphi\ \text{sin}\ \theta} & & {\rho\ \text{cos}\ \varphi\ \text{cos}\ \theta} \\ {\text{sin}\ \varphi\ \text{sin}\ \theta} & & {\rho\ \text{sin}\ \varphi\ \text{cos}\ \theta} & & {\rho\ \text{cos}\ \varphi\ \text{sin}\ \theta} \\ 0 & & 0 & & {- \rho\ \text{sin}\ \varphi} \end{array} \right|.$$

Expanding the determinant with respect to the third row:

$$\begin{matrix} \\ \\ {= \text{cos}\ \varphi\left| \begin{array}{lll} {- \rho\ \text{sin}\ \varphi\ \text{sin}\ \theta} & & {\rho\ \text{cos}\ \varphi\ \text{cos}\ \theta} \\ {\rho\ \text{sin}\ \varphi\ \text{sin}\ \theta} & & {\rho\ \text{cos}\ \varphi\ \text{sin}\ \theta} \end{array} \right| - \rho\ \text{sin}\ \varphi\left| \begin{array}{lll} {\text{sin}\ \varphi\ \text{cos}\ \theta} & & {- \rho\ \text{sin}\ \varphi\ \text{sin}\ \theta} \\ {\text{sin}\ \varphi\ \text{sin}\ \theta} & & {\rho\ \text{sin}\ \varphi\ \text{cos}\ \theta} \end{array} \right|} \\ {= \text{cos}\ \varphi\left( {\text{−}\rho^{2}\text{sin}\ \varphi\ \text{cos}\ \varphi\ \text{sin}^{2}\theta - \rho^{2}\text{sin}\ \varphi\ \text{cos}\ \varphi\ \text{cos}^{2}\theta} \right)} \\ {\quad - \rho\ \text{sin}\ \varphi\left( {\rho\ \text{sin}^{2}\varphi\ \text{cos}^{2}\theta + \rho\ \text{sin}^{2}\varphi\ \text{sin}^{2}\theta} \right)} \\ {= \text{−}\rho^{2}\text{sin}\ \varphi\ \text{cos}^{2}\varphi\left( {\text{sin}^{2}\theta + \text{cos}^{2}\theta} \right) - \rho^{2}\text{sin}\ \varphi\ \text{sin}^{2}\varphi\left( {\text{sin}^{2}\theta + \text{cos}^{2}\theta} \right)} \\ {= \text{−}\rho^{2}\text{sin}\ \varphi\ \text{cos}^{2}\varphi - \rho^{2}\text{sin}\ \varphi\ \text{sin}^{2}\varphi} \\ {= \text{−}\rho^{2}\text{sin}\ \varphi\left( {\text{cos}^{2}\varphi + \text{sin}^{2}\varphi} \right) = \text{−}\rho^{2}\text{sin}\ \varphi.} \end{matrix}$$

Since $0 \leq \varphi \leq \pi,$ we must have $\text{sin}\ \varphi \geq 0.$ Thus $\left| {J\left( {\rho,\theta,\varphi} \right)} \right| = \left| {\text{−}\rho^{2}\text{sin}\ \varphi} \right| = \rho^{2}\text{sin}\ \varphi.$

Then the triple integral becomes

$${\iiint\limits_{D}{f\left( {x,y,z} \right)}}dV = {\iiint\limits_{G}{f\left( {\rho\ \text{sin}\ \varphi\ \text{cos}\ \theta,\rho\ \text{sin}\ \varphi\ \text{sin}\ \theta,\rho\ \text{cos}\ \varphi} \right)}}\rho^{2}\text{sin}\ \varphi\ d\rho\ d\varphi\ d\theta.$$

Solution 解

As before, some kind of sketch of the region $G$ in $xyz\text{-space}$ over which we have to perform the integration can help identify the region $D$ in $uvw\text{-space}$ (Figure 5.83). Clearly $G$ in $xyz\text{-space}$ is bounded by the planes $x = y\text{/}2,x = \left( {y\text{/}2} \right) + 1,y = 0,$ $y = 4,$ $z = 0,\text{and}\ z = 4.$ We also know that we have to use $u = \left( {2x - y} \right)\text{/}2,v = y\text{/}2,\text{and}\ w = z\text{/}3$ for the transformations. We need to solve for $x,y,\text{and}\ z.$ Here we find that $x = u + v,$ $y = 2v,$ and $z = 3w.$

Using elementary algebra, we can find the corresponding surfaces for the region $G$ and the limits of integration in $uvw\text{-space}\text{.}$ It is convenient to list these equations in a table.

| Equations in $xyz$ for the region $D$ | Corresponding equations in $uvw$ for the region $G$ | Limits for the integration in $uvw$ |

|---------------------------------------|-----------------------------------------------------|-------------------------------------|

| $x = y\text{/}2$ | $u + v = 2v\text{/}2 = v$ | $u = 0$ |

| $x = y\text{/}2$ | $u + v = \left( {2v\text{/}2} \right) + 1 = v + 1$ | $u = 1$ |

| $y = 0$ | $2v = 0$ | $v = 0$ |

| $y = 4$ | $2v = 4$ | $v = 2$ |

| $z = 0$ | $3w = 0$ | $w = 0$ |

| $z = 3$ | $3w = 3$ | $w = 1$ |

Now we can calculate the Jacobian for the transformation:

$$J(u,v,w) = \left| \begin{array}{lclcl} \frac{\partial x}{\partial u} & & \frac{\partial x}{\partial v} & & \frac{\partial x}{\partial w} \\ \frac{\partial y}{\partial u} & & \frac{\partial y}{\partial v} & & \frac{\partial y}{\partial w} \\ \frac{\partial z}{\partial u} & & \frac{\partial z}{\partial v} & & \frac{\partial z}{\partial w} \end{array} \right| = \left| \begin{array}{lclcl} 1 & & 1 & & 0 \\ 0 & & 2 & & 0 \\ 0 & & 0 & & 3 \end{array} \right| = 6.$$

The function to be integrated becomes

$$f\left( {x,y,z} \right) = x + \frac{z}{3} = u + v + \frac{3w}{3} = u + v + w.$$

We are now ready to put everything together and complete the problem.

$$\begin{array}{l} \\ \\ \\ \\ {\quad{\int\limits_{0}^{3}\ {\int\limits_{0}^{4}\ {\int\limits_{y\text{/}2}^{{({y\text{/}2})} + 1}\left( {x + \frac{z}{3}} \right)}}}dx\ dy\ dz} \\ \\ {= {\int\limits_{0}^{1}\ {\int\limits_{0}^{2}\ {\int\limits_{0}^{1}\left( {u + v + w} \right)}}}\left| {J\left( {u,v,w} \right)} \right|du\ dv\ dw = {\int\limits_{0}^{1}\ {\int\limits_{0}^{2}\ {\int\limits_{0}^{1}\left( {u + v + w} \right)}}}|6|du\ dv\ dw} \\ {= 6{\int\limits_{0}^{1}\ {\int\limits_{0}^{2}\ {\int\limits_{0}^{1}\left( {u + v + w} \right)}}}du\ dv\ dw = 6{\int\limits_{0}^{1}\ {\int\limits_{0}^{2}\left\lbrack {\frac{u^{2}}{2} + vu + wu} \right\rbrack}}_{0}^{1}dv\ dw} \\ {= 6{\int\limits_{0}^{1}\ {\int\limits_{0}^{2}\left( {\frac{1}{2} + v + w} \right)}}dv\ dw = 6{\int\limits_{0}^{1}\left\lbrack {\frac{1}{2}v + \frac{v^{2}}{2} + wv} \right\rbrack}_{0}^{2}dw} \\ {= 6{\int\limits_{0}^{1}\left( {3 + 2w} \right)}dw = 6\left\lbrack {3w + w^{2}} \right\rbrack_{0}^{1} = 24.} \end{array}$$

Let $D$ be the region in $xyz\text{-space}$ defined by $1 \leq x \leq 2,0 \leq xy \leq 2,\text{and}\ 0 \leq z \leq 1.$

Evaluate ${\iiint\limits_{D}\left( {x^{2}y + 3xyz} \right)}dx\ dy\ dz$ by using the transformation $u = x,v = xy,$ and $w = 3z.$

Section 5.7 Exercises 5.7 节习题

In the following exercises, the function $T:S\rightarrow R,T\left( {u,v} \right) = \left( {x,y} \right)$ on the region $S = \left\{ {\left. \left( {u,v} \right) \right|0 \leq u \leq 1,0 \leq v \leq 1} \right\}$ bounded by the unit square is given, where $R \subset \mathbb{R}^{2}$ is the image of $S$ under $T.$

1. Justify that the function $T$ is a $C^{1}$ transformation.

2. Find the images of the vertices of the unit square $S$ through the function $T.$

3. Determine the image $R$ of the unit square $S$ and graph it.

356\.

$x = 2u,y = 3v$

357\.

$x = \frac{u}{2},y = \frac{v}{3}$

358\.

$x = u - v,y = u + v$

359\.

$x = 2u - v,y = u + 2v$

360\.

$x = u^{2},y = v^{2}$

361\.

$x = u^{3},y = v^{3}$

In the following exercises, determine whether the transformations $T:S\rightarrow R$ are one-to-one or not.

362\.

$x = u^{2},y = v^{2},\text{where}\ S$ is the rectangle of vertices $\left( {-1,0} \right),\left( {1,0} \right),\left( {1,1} \right),\text{and}\ \left( {-1,1} \right).$

363\.

$x = u^{4},y = u^{2} + v,\text{where}\ S$ is the triangle of vertices $\left( {-2,0} \right),\left( {2,0} \right),\text{and}\ \left( {0,2} \right).$

364\.

$x = 2u,y = 3v,\text{where}\ S$ is the square of vertices $\left( {-1,1} \right),\left( {-1,-1} \right),\left( {1,-1} \right),\text{and}\ \left( {1,1} \right).$

365\.

$T\left( {u,v} \right) = \left( {2u - v,u} \right),$ where $S$ is the triangle of vertices $\left( {-1,1} \right),\left( {-1,-1} \right),\text{and}\ \left( {1,-1} \right).$

366\.

$x = u + v + w,y = u + v,z = w,$ where $S = R = \mathbb{R}^{3}.$

367\.

$x = u^{2} + v + w,y = u^{2} + v,z = w,$ where $S = R = \mathbb{R}^{3}.$

In the following exercises, the transformations $T:S\rightarrow R$ are one-to-one. Find their related inverse transformations $T^{-1}:R\rightarrow S.$

368\.

$x = 4u,y = 5v,$ where $S = R = \mathbb{R}^{2}.$

369\.

$x = u + 2v,y = \text{−}u + v,$ where $S = R = \mathbb{R}^{2}.$

370\.

$x = e^{2u + v},y = e^{u - v},$ where $S = \mathbb{R}^{2}$ and $R = \left\{ {\left. \left( {x,y} \right) \right|x > 0,y > 0} \right\}$

371\.

$x = \text{ln}\ u,y = \text{ln}\left( {uv} \right),$ where $S = \left\{ {\left. \left( {u,v} \right) \right|u > 0,v > 0} \right\}$ and $R = \mathbb{R}^{2}.$

372\.

$x = u + v + w,y = 3v,z = 2w,$ where $S = R = \mathbb{R}^{3}.$

373\.

$x = u + v,y = v + w,z = u + w,$ where $S = R = \mathbb{R}^{3}.$

In the following exercises, the transformation $T:S\rightarrow R,T\left( {u,v} \right) = \left( {x,y} \right)$ and the region $R \subset \mathbb{R}^{2}$ are given. Find the region $S \subset \mathbb{R}^{2}.$

374\.

$x = au,y = bv,R = \left\{ {\left. \left( {x,y} \right) \right|x^{2} + y^{2} \leq a^{2}b^{2}} \right\},$ where $a,b > 0$

375\.

$x = au,y = bv,R = \left\{ {\left. \left( {x,y} \right) \right|\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} \leq 1} \right\},$ where $a,b > 0$

376\.

$x = \frac{u}{a},y = \frac{v}{b},z = \frac{w}{c},$ $R = \left\{ {\left. \left( {x,y} \right) \right|x^{2} + y^{2} + z^{2} \leq 1} \right\},$ where $a,b,c > 0$

377\.

$x = au,y = bv,z = cw,R = \left\{ {\left. \left( {x,y} \right) \right|\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} - \frac{z^{2}}{c^{2}} \leq 1,z > 0} \right\},$ where $a,b,c > 0$

In the following exercises, find the Jacobian $J$ of the transformation.

378\.

$x = u + 2v,y = \text{−}u + v$

379\.

$x = \frac{u^{3}}{2},y = \frac{v}{u^{2}}$

380\.

$x = e^{2u - v},y = e^{u + v}$

381\.

$x = ue^{v},y = e^{\text{−}v}$

382\.

$x = u\ \text{cos}\left( e^{v} \right),y = u\ \text{sin}\left( e^{v} \right)$

383\.

$x = v\ \text{sin}\left( u^{2} \right),y = v\ \text{cos}\left( u^{2} \right)$

384\.

$x = u\ \text{cosh}\ v,y = u\ \text{sinh}\ v,z = w$

385\.

$x = v\ \text{cosh}\left( \frac{1}{u} \right),y = v\ \text{sinh}\left( \frac{1}{u} \right),z = u + w^{2}$

386\.

$x = u + v,y = v + w,z = u$

387\.

$x = u - v,y = u + v,z = u + v + w$

388\.

The triangular region $R$ with the vertices $\left( {0,0} \right),\left( {1,1} \right),\text{and}\ \left( {1,2} \right)$ is shown in the following figure.

1. Find a transformation $T:S\rightarrow R,$ $T\left( {u,v} \right) = \left( {x,y} \right) = \left( {au + bv,cu + dv} \right),$ where $a,b,c,$ and $d$ are real numbers with $ad - bc \neq 0$ such that $T^{-1}\left( {0,0} \right) = \left( {0,0} \right),T^{-1}\left( {1,1} \right) = \left( {1,0} \right),$ and $T^{-1}\left( {1,2} \right) = \left( {0,1} \right).$

2. Use the transformation $T$ to find the area $A(R)$ of the region $R.$

389\.

The triangular region $R$ with the vertices $\left( {0,0} \right),\left( {2,0} \right),\text{and}\ \left( {1,3} \right)$ is shown in the following figure.

1. Find a transformation $T:S\rightarrow R,$ $T\left( {u,v} \right) = \left( {x,y} \right) = \left( {au + bv,cu + dv} \right),$ where $a,b,c$ and $d$ are real numbers with $ad - bc \neq 0$ such that $T^{-1}\left( {0,0} \right) = \left( {0,0} \right),$ $T^{-1}\left( {2,0} \right) = \left( {1,0} \right),$ and $T^{-1}\left( {1,3} \right) = \left( {0,1} \right).$

2. Use the transformation $T$ to find the area $A(R)$ of the region $R.$

In the following exercises, use the transformation $u = y - x,v = y,$ to evaluate the integrals on the parallelogram $R$ of vertices $\left( {0,0} \right),\left( {1,0} \right),\left( {2,1} \right),\text{and}\ \left( {1,1} \right)$ shown in the following figure.

390\.

${\iint\limits_{R}\left( {y - x} \right)}dA$

391\.

${\iint\limits_{R}\left( {y^{2} - xy} \right)}dA$

In the following exercises, use the transformation $y - x = u,x + y = v$ to evaluate the integrals on the square $R$ determined by the lines $y = x,y = \text{−}x + 2,y = x + 2,$ and $y = \text{−}x$ shown in the following figure.

392\.

$\iint\limits_{R}{e^{x + y}dA}$

393\.

${\iint\limits_{R}{\text{sin}\left( {x - y} \right)}}dA$

In the following exercises, use the transformation $x = u,5y = v$ to evaluate the integrals on the region $R$ bounded by the ellipse $x^{2} + 25y^{2} = 1$ shown in the following figure.

394\.

$\iint\limits_{R}{\sqrt{x^{2} + 25y^{2}}\ dA}$

395\.

${\iint\limits_{R}\left( {x^{2} + 25y^{2}} \right)}^{2}dA$

In the following exercises, use the transformation $u = x + y,v = x - y$ to evaluate the integrals on the trapezoidal region $R$ determined by the points $\left( {1,0} \right),\left( {2,0} \right),\left( {0,2}\text{)} \right.,\text{and}\ \left( {0,1} \right)$ shown in the following figure.

396\.

$\iint\limits_{R}{\left( {x^{2} - 2xy + y^{2}} \right)e^{x + y}dA}$

397\.

${\iint\limits_{R}\left( {x^{3} + 3x^{2}y + 3xy^{2} + y^{3}} \right)}dA$

398\.

The circular annulus sector $R$ bounded by the circles $4x^{2} + 4y^{2} = 1$ and $9x^{2} + 9y^{2} = 64,$ the line $x = y\sqrt{3},$ and the $y\text{-axis}$ is shown in the following figure. Find a transformation $T$ from a rectangular region $S$ in the $r\theta\text{-plane}$ to the region $R$ in the $xy\text{-plane.}$ Graph $S.$

399\.

The solid $R$ bounded by the circular cylinder $x^{2} + y^{2} = 9$ and the planes $z = 0,z = 1,$ $x = 0,\text{and}\ y = 0$ is shown in the following figure. Find a transformation $T$ from a cylindrical box $S$ in $r\theta z\text{-space}$ to the solid $R$ in $xyz\text{-space}.$

400\.

Show that ${\iint\limits_{R}{f\left( \sqrt{\frac{x^{2}}{3} + \frac{y^{2}}{3}} \right)}}dA = 2\pi\sqrt{15}{\int\limits_{0}^{1}{f(\rho)}}\rho\ d\rho,$ where $f$ is a continuous function on $\left\lbrack {0,1} \right\rbrack$ and $R$ is the region bounded by the ellipse $5x^{2} + 3y^{2} = 15.$

401\.

Show that ${\iiint\limits_{R}{f\left( \sqrt{16x^{2} + 4y^{2} + z^{2}} \right)}}dV = \frac{\pi}{2}{\int\limits_{0}^{1}{f(\rho)}}\rho^{2}d\rho,$ where $f$ is a continuous function on $\left\lbrack {0,1} \right\rbrack$ and $R$ is the region bounded by the ellipsoid $16x^{2} + 4y^{2} + z^{2} = 1.$

402\.

\[T\] Find the area of the region bounded by the curves $xy = 1,xy = 3,y = 2x,$ and $y = 3x$ by using the transformation $u = xy$ and $v = \frac{y}{x}.$ Use a computer algebra system (CAS) to graph the boundary curves of the region $R.$

403\.

\[T\] Find the area of the region bounded by the curves $xy = 2,xy = 3,y = x,$ and $y = 2x$ by using the transformation $u = xy$ and $v = \frac{y}{x}.$

404\.

Evaluate the triple integral $\int\limits_{0}^{1}\ {\int\limits_{1}^{2}\ {\int\limits_{z}^{z + 1}{\left( {y + 1} \right)dx\ dy\ dz}}}$ by using the transformation $u = x - z,$ $v = 3y,\text{and}\ w = \frac{z}{2}.$

405\.

Evaluate the triple integral $\int\limits_{0}^{2}\ {\int\limits_{4}^{6}\ {\int\limits_{3z}^{3z + 2}{\left( {5 - 4y} \right)dx\ dz\ dy}}}$ by using the transformation $u = x - 3z,v = 4y,\text{and}\ w = z.$

406\.

A transformation $T:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2},T\left( {u,v} \right) = \left( {x,y} \right)$ of the form $x = au + bv,y = cu + dv,$ where $a,b,c,\text{and}\ d$ are real numbers, is called linear. Show that a linear transformation for which $ad - bc \neq 0$ maps parallelograms to parallelograms.

407\.

The transformation $T_{\theta}:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2},T_{\theta}\left( {u,v} \right) = \left( {x,y} \right),$ where $x = u\ \text{cos}\ \theta - v\ \text{sin}\ \theta,$ $y = u\ \text{sin}\ \theta + v\ \text{cos}\ \theta,$ is called a rotation of angle $\theta.$ Show that the inverse transformation of $T_{\theta}$ satisfies $T_{\theta}{}^{-1} = T_{\text{−}\theta},$ where $T_{\text{−}\theta}$ is the rotation of angle $\text{−}\theta.$

408\.

\[T\] The transformations $T_{i}:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2},$ $i = 1\text{,…,}\ 4,$ defined by $T_{1}\left( {u,v} \right) = \left( {u,\text{−}v} \right),$ $T_{2}\left( {u,v} \right) = \left( {\text{−}u,v} \right),T_{3}\left( {u,v} \right) = \left( {\text{−}u,\text{−}v} \right),$ and $T_{4}\left( {u,v} \right) = \left( {v,u} \right)$ are called reflections about the $x\text{-axis},y\text{-axis},$ origin, and the line $y = x,$ respectively.

1. Find the image of the region $S = \left\{ {\left. \left( {u,v} \right) \right|u^{2} + v^{2} - 2u - 4v + 1 \leq 0} \right\}$ in the $xy\text{-plane}$ through the transformation $T_{1} \circ T_{2} \circ T_{3} \circ T_{4}.$

2. Use a CAS to graph the image of $S.$

3. Evaluate the integral $\iint\limits_{S}{\text{sin}(u^{2})du\ dv}$ by using a CAS. Round your answer to two decimal places.

409\.

\[T\] The transformation $T_{k,1,1}:\mathbb{R}^{3}\rightarrow\mathbb{R}^{3},T_{k,1,1}\left( {u,v,w} \right) = \left( {x,y,z} \right)$ of the form $x = ku,$ $y = v,z = w,$ where $k \neq 1$ is a positive real number, is called a stretch if $k > 1$ and a compression if $0 < k < 1$ in the $x\text{-direction}\text{.}$ Use a CAS to evaluate the integral ${\iiint\limits_{S}e^{\text{−}{({4x^{2} + 9y^{2} + 25z^{2}})}\ dy\ dz}$ on the solid $S = \left\{ {\left. \left( {x,y,z} \right) \right|4x^{2} + 9y^{2} + 25z^{2} \leq 1} \right\}$ by considering the compression $T_{2,3,5}\left( {u,v,w} \right) = \left( {x,y,z} \right)$ defined by $x = \frac{u}{2},y = \frac{v}{3},$ and $z = \frac{w}{5}.$ Round your answer to four decimal places.

410\.

\[T\] The transformation $T_{a,0}:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2},T_{a,0}\left( {u,v} \right) = \left( {u + av,v} \right),$ where $a \neq 0$ is a real number, is called a shear in the $x\text{-direction}\text{.}$ The transformation, $T_{0,b}:\mathbb{R}^{2}\rightarrow\mathbb{R}^{2},T_{0,b}\left( {u,v} \right) = \left( {u,bu + v} \right),$ where $b \neq 0$ is a real number, is called a shear in the $y\text{-direction}\text{.}$

1. Find transformations $T_{0,2} \circ T_{3,0}.$

2. Find the image $R$ of the trapezoidal region $S$ bounded by $u = 0,v = 0,v = 1,$ and $v = 2 - u$ through the transformation $T_{0,2} \circ T_{3,0}.$

3. Use a CAS to graph the image $R$ in the $xy\text{-plane}\text{.}$

4. Find the area of the region $R$ by using the area of region $S.$

411\.

Use the transformation, $x = au,y = av,z = cw$ and spherical coordinates to show that the volume of a region bounded by the spheroid $\frac{x^{2} + y^{2}}{a^{2}} + \frac{z^{2}}{c^{2}} = 1$ is $\frac{4\pi a^{2}c}{3}.$

412\.

Find the volume of a football whose shape is a spheroid $\frac{x^{2} + y^{2}}{a^{2}} + \frac{z^{2}}{c^{2}} = 1$ whose length from tip to tip is $11$ inches and circumference at the center is $22$ inches. Round your answer to two decimal places.

413\.

\[T\] Lamé ovals (or superellipses) are plane curves of equations $\left( \frac{x}{a} \right)^{n} + \left( \frac{y}{b} \right)^{n} = 1,$ where *a*, *b*, and *n* are positive real numbers.

1. Use a CAS to graph the regions $R$ bounded by Lamé ovals for $a = 1,b = 2,n = 4$ and $n = 6,$ respectively.

2. Find the transformations that map the region $R$ bounded by the Lamé oval $x^{4} + y^{4} = 1,$ also called a squircle and graphed in the following figure, into the unit disk.

3. Use a CAS to find an approximation of the area $A(R)$ of the region $R$ bounded by $x^{4} + y^{4} = 1.$ Round your answer to two decimal places.

414\.

\[T\] Lamé ovals (or superellipses) are plane curves of equations $\left( \frac{x}{a} \right)^{n} + \left( \frac{y}{b} \right)^{n} = 1,$ where *a*, *b*, and *n* are positive real numbers.

415\.

\[T\] Lamé ovals have been consistently used by designers and architects. For instance, Gerald Robinson, a Canadian architect, has designed a parking garage in a shopping center in Peterborough, Ontario, in the shape of a superellipse of the equation $\left( \frac{x}{a} \right)^{n} + \left( \frac{y}{b} \right)^{n} = 1$ with $\frac{a}{b} = \frac{9}{7}$ and $n = e.$ Use a CAS to find an approximation of the area of the parking garage in the case $a = 900$ yards, $b = 700$ yards, and $n = 2.72$.

---

Key Terms 关键术语

double integral

of the function $f(x,y)$ over the region $R$ in the $xy$-plane is defined as the limit of a double Riemann sum, ${\iint\limits_{R}{f(x,y)dA}} = \underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f(x_{ij}^{*},y_{ij}^{*})\text{Δ}A.}}}$

double Riemann sum

of the function $f(x,y)$ over a rectangular region $R$ is $\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f(x_{ij}^{*},y_{ij}^{*})\text{Δ}A}}$ where $R$ is divided into smaller subrectangles $R_{ij}$ and $(x_{ij}^{*},y_{ij}^{*})$ is an arbitrary point in $R_{ij}$

Fubini’s theorem

if $f(x,y)$ is a function of two variables that is continuous over a rectangular region $R = \left\{ (x,y) \in \mathbb{R}^{2}\left| \left. a \leq x \leq b,c \leq y \leq d \right\} \right., \right.$ then the double integral of $f$ over the region equals an iterated integral, ${\iint\limits_{R}{f(x,y)dy\ dx}} = {\int_{a}^{b}{\int_{c}^{d}{f(x,y)dx\ dy}}} = {\int_{c}^{d}{\int_{a}^{b}{f(x,y)dx\ dy}}}$

improper double integral

a double integral over an unbounded region or of an unbounded function

iterated integral

for a function $f(x,y)$ over the region $R$ is

1. ${\int_{a}^{b}{\int_{c}^{d}{f(x,y)dx\ dy}}} = {\int_{a}^{b}\left\lbrack {\int_{c}^{d}{f(x,y)dy}} \right\rbrack}dx,$

2. ${\int_{c}^{d}{\int_{b}^{a}{f(x,y)dx\ dy}}} = {\int_{c}^{d}\left\lbrack {\int_{a}^{b}{f(x,y)dx}} \right\rbrack}dy,$

where $a,b,c,$ and $d$ are any real numbers and $R = \lbrack a,b\rbrack\ \times \ \lbrack c,d\rbrack$

Jacobian

the Jacobian $J\left( {u,v} \right)$ in two variables is a $2\ \times \ 2$ determinant:

$$J\left( {u,v} \right) = \left| \begin{array}{lll} \frac{\partial x}{\partial u} & & \frac{\partial y}{\partial u} \\ \frac{\partial x}{\partial v} & & \frac{\partial y}{\partial v} \end{array} \right|;$$

the Jacobian $J\left( {u,v,w} \right)$ in three variables is a $3\ \times \ 3$ determinant:

$$J(u,v,w) = \left| \begin{array}{lclcl} \frac{\partial x}{\partial u} & & \frac{\partial y}{\partial u} & & \frac{\partial z}{\partial u} \\ \frac{\partial x}{\partial v} & & \frac{\partial y}{\partial v} & & \frac{\partial z}{\partial v} \\ \frac{\partial x}{\partial w} & & \frac{\partial y}{\partial w} & & \frac{\partial z}{\partial w} \end{array} \right|$$

one-to-one transformation

a transformation $T:G\rightarrow R$ defined as $T\left( {u,v} \right) = \left( {x,y} \right)$ is said to be one-to-one if no two points map to the same image point

planar transformation

a function $T$ that transforms a region $G$ in one plane into a region $R$ in another plane by a change of variables

polar rectangle

the region enclosed between the circles $r = a$ and $r = b$ and the angles $\theta = \alpha$ and $\theta = \beta;$ it is described as $R = \left\{ {\left. \left( {r,\theta} \right) \right|a \leq r \leq b,\alpha \leq \theta \leq \beta} \right\}$

radius of gyration

the distance between the rotational axis of the object and the point where the entire mass of the object can be concentrated and have the same moment of inertia

transformation

a function that transforms a region $G$ in one plane into a region $R$ in another plane by a change of variables

triple integral

the triple integral of a continuous function $f\left( {x,y,z} \right)$ over a rectangular solid box $B$ is the limit of a Riemann sum for a function of three variables, if this limit exists

triple integral in cylindrical coordinates

the limit of a triple Riemann sum, provided the following limit exists:

$$\underset{l,m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{l}{\sum\limits_{j = 1}^{m}{\sum\limits_{k = 1}^{n}{f(r_{ijk}^{*},\theta_{ijk}^{*},z_{ijk}^{*})r_{ijk}^{*}\text{Δ}r\text{Δ}\theta\text{Δ}z}}}}$$

triple integral in spherical coordinates

the limit of a triple Riemann sum, provided the following limit exists:

$$\underset{l,m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{l}{\sum\limits_{j = 1}^{m}{\sum\limits_{k = 1}^{n}{f(\rho_{ijk}^{*},\theta_{ijk}^{*},\varphi_{ijk}^{*}){(\rho_{ijk}^{*})}^{2}\text{sin}\ \varphi\text{Δ}\rho\text{Δ}\theta\text{Δ}\varphi}}}}$$

Type I

a region $D$ in the $xy$-plane is Type I if it lies between two vertical lines and the graphs of two continuous functions $g_{1}(x)$ and $g_{2}(x)$

Type II

a region $D$ in the $xy$-plane is Type II if it lies between two horizontal lines and the graphs of two continuous functions $h_{1}(y)\ \text{and}\ h_{2}(y)$

Key Equations 关键公式

| | |

|---------------------------------------------|------------------------------------------------------------------------------------------------------------------------------------------------------------------------|

| Iterated integral over a Type I region | ${\iint\limits_{D}{f(x,y)dA}} = {\iint\limits_{D}{f(x,y)dy\ dx}} = {\int\limits_{a}^{b}{\left\lbrack {\int\limits_{g_{1}(x)}^{g_{2}(x)}{f(x,y)dy}} \right\rbrack dx}}$ |

| Iterated integral over a Type II region | ${\iint\limits_{D}{f(x,y)dA}} = {\iint\limits_{D}{f(x,y)dx\ dy}} = {\int\limits_{c}^{d}{\left\lbrack {\int\limits_{h_{1}(y)}^{h_{2}(y)}{f(x,y)dx}} \right\rbrack dy}}$ |

| | |

|---------------------------------------------------------|-----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|

| Double integral over a polar rectangular region $R$ | ${\iint\limits_{R}{f\left( {r,\theta} \right)}}dA = \underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f\left( {r_{ij}*,\theta_{ij}*} \right)}}}\Delta A = \underset{m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{m}{\sum\limits_{j = 1}^{n}{f\left( {r_{ij}*,\theta_{ij}*} \right)}}}r_{ij}*\Delta r\Delta\theta$ |

| Double integral over a general polar region | ${\iint\limits_{D}{f\left( {r,\theta} \right)}}r\ dr\ d\theta = {\int\limits_{\theta = \alpha}^{\theta = \beta}\ {\int\limits_{r = h_{1}{(\theta)}}^{r = h_{2}{(\theta)}}{f\left( {r,\theta} \right)}}}r\ dr\ d\theta$ |

| | |

|---------------------|---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|

| Triple integral | $\underset{l,m,n\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{l}{\sum\limits_{j = 1}^{m}{\sum\limits_{k = 1}^{n}{f(x_{ijk}^{*},y_{ijk}^{*},z_{ijk}^{*})\text{Δ}x\text{Δ}y\text{Δ}z}}}} = {\iiint\limits_{B}{f(x,y,z)dV}}$ |

| | |

|------------------------------------------------|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|

| Triple integral in cylindrical coordinates | ${\iiint\limits_{B}{g\left( {x,y,z} \right)dV}} = {\iiint\limits_{B}{g\left( {r\ \text{cos}\ \theta,r\ \text{sin}\ \theta,z} \right)r\ dr\ d\theta\ dz = {\iiint\limits_{B}{f\left( {r,\theta,z} \right)}}}}r\ dr\ d\theta\ dz$ |

| Triple integral in spherical coordinates | ${\iiint\limits_{B}{f\left( {\rho,\theta,\varphi} \right)}}\rho^{2}\text{sin}\ \varphi\ d\rho\ d\varphi\ d\theta = {\int\limits_{\varphi = \gamma}^{\varphi = \psi}\ {\int\limits_{\theta = \alpha}^{\theta = \beta}\ {\int\limits_{\rho = a}^{\rho = b}{f\left( {\rho,\theta,\varphi} \right)}}}}\rho^{2}\text{sin}\ \varphi\ d\rho\ d\varphi\ d\theta$ |

| | |

|--------------------------------|--------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------|

| Mass of a lamina | $m = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}m_{ij}}} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}{\rho(x_{ij}^{*},y_{ij}^{*})\text{Δ}A = {\iint\limits_{R}{\rho(x,y)dA}}}}}$ |

| **Moment about the *x*-axis** | $M_{x} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}{\left( y_{ij}^{*} \right)m_{ij}}}} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}{\left( y_{ij}^{*} \right)\rho(x_{ij}^{*},y_{ij}^{*})\text{Δ}A = {\iint\limits_{R}{y\rho(x,y)dA}}}}}$ |

| **Moment about the *y*-axis** | $M_{y} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}{\left( x_{ij}^{*} \right)m_{ij}}}} = \underset{k,l\rightarrow\infty}{\text{lim}}{\sum\limits_{i = 1}^{k}{\sum\limits_{j = 1}^{l}{\left( x_{ij}^{*} \right)\rho(x_{ij}^{*},y_{ij}^{*})\text{Δ}A = {\iint\limits_{R}{x\rho(x,y)dA}}}}}$ |

| Center of mass of a lamina | $\overset{\text{−}}{x} = \frac{M_{y}}{m} = \frac{\iint\limits_{R}{x\rho\left( {x,y} \right)dA}}{\iint\limits_{R}{\rho\left( {x,y} \right)dA}}$ and $\overset{\text{−}}{y} = \frac{M_{x}}{m} = \frac{\iint\limits_{R}{y\rho\left( {x,y} \right)dA}}{\iint\limits_{R}{\rho\left( {x,y} \right)dA}}$ |

Key Concepts 关键概念

5.1 Double Integrals over Rectangular Regions 5.1 矩形区域上的二重积分

• We can use a double Riemann sum to approximate the volume of a solid bounded above by a function of two variables over a rectangular region. By taking the limit, this becomes a double integral representing the volume of the solid.
• Properties of double integral are useful to simplify computation and find bounds on their values.
• We can use Fubini's theorem to write and evaluate a double integral as an iterated integral.
• Double integrals are used to calculate the area of a region, the volume under a surface, and the average value of a function of two variables over a rectangular region.

• 我们可以用二重黎曼和来近似由二元函数在矩形区域上的图像为上界所围立体的体积；取极限后，它就成为表示该立体体积的二重积分。
• 二重积分的性质有助于简化计算并求出其值的界。
• 我们可以利用富比尼定理，把二重积分写成累次积分并求其值。
• 二重积分可用于计算区域的面积、曲面下的体积，以及二元函数在矩形区域上的平均值。

5.2 Double Integrals over General Regions 5.2 一般区域上的二重积分

• A general bounded region $D$ on the plane is a region that can be enclosed inside a rectangular region. We can use this idea to define a double integral over a general bounded region.
• To evaluate an iterated integral of a function over a general nonrectangular region, we sketch the region and express it as a Type I or as a Type II region or as a union of several Type I or Type II regions that overlap only on their boundaries.
• We can use double integrals to find volumes, areas, and average values of a function over general regions, similarly to calculations over rectangular regions.
• We can use Fubini's theorem for improper integrals to evaluate some types of improper integrals.

• 平面上的一个一般有界区域 $D$ 是指可被包围在某个矩形区域之内的区域。我们可借此定义一般有界区域上的二重积分。
• 为计算函数在一般非矩形区域上的累次积分，我们先画出该区域，并将其表示为Ⅰ型区域或Ⅱ型区域，或表示为若干仅在边界上相交的Ⅰ型、Ⅱ型区域的并。
• 与矩形区域上的情形类似，我们可用二重积分求函数在一般区域上的体积、面积与平均值。
• 对于某些类型的反常积分，我们可以利用关于反常积分的富比尼定理来计算。

5.3 Double Integrals in Polar Coordinates 5.3 极坐标下的二重积分

• To apply a double integral to a situation with circular symmetry, it is often convenient to use a double integral in polar coordinates. We can apply these double integrals over a polar rectangular region or a general polar region, using an iterated integral similar to those used with rectangular double integrals.
• The area $dA$ in polar coordinates becomes $r\ dr\ d\theta.$
• Use $x = r\ \text{cos}\ \theta,$ $y = r\ \text{sin}\ \theta,$ and $dA = r\ dr\ d\theta$ to convert an integral in rectangular coordinates to an integral in polar coordinates.
• Use $r^{2} = x^{2} + y^{2}$ and $\theta = \text{tan}^{-1}\left( \frac{y}{x} \right)$ to convert an integral in polar coordinates to an integral in rectangular coordinates, if needed.
• To find the volume in polar coordinates bounded above by a surface $z = f\left( {r,\theta} \right)$ over a region on the $xy$-plane, use a double integral in polar coordinates.

• 在具有圆对称性的情形中应用二重积分时，通常宜使用极坐标下的二重积分。我们可在极矩形区域或一般极坐标区域上应用这些二重积分，所用累次积分与矩形区域上的二重积分类似。
• 在极坐标下，面积元 $dA$ 变为 $r\ dr\ d\theta$。
• 利用 $x = r\ \text{cos}\ \theta,$ $y = r\ \text{sin}\ \theta,$ 以及 $dA = r\ dr\ d\theta$，把直角坐标下的积分转换为极坐标下的积分。
• 如有需要，利用 $r^{2} = x^{2} + y^{2}$ 与 $\theta = \text{tan}^{-1}\left( \frac{y}{x} \right)$，把极坐标下的积分转换回直角坐标下的积分。
• 为求 $xy$ 平面上某区域上方、以曲面 $z = f\left( {r,\theta} \right)$ 为上界的极坐标体积，可使用极坐标下的二重积分。

5.4 Triple Integrals 5.4 三重积分

• To compute a triple integral we use Fubini's theorem, which states that if $f\left( {x,y,z} \right)$ is continuous on a rectangular box $B = \left\lbrack {a,b} \right\rbrack\ \times \ \left\lbrack {c,d} \right\rbrack\ \times \ \left\lbrack {e,f} \right\rbrack,$ then

$$\iiint\limits_{B}{f\left( {x,y,z} \right)dV = {\int\limits_{e}^{f}\ {\int\limits_{c}^{d}\ {\int\limits_{a}^{b}{f\left( {x,y,z} \right)dx\ dy\ dz}}}}}$$

and is also equal to any of the other five possible orderings for the iterated triple integral.

• To compute the volume of a general solid bounded region $E$ we use the triple integral

$$V(E) = {\iiint\limits_{E}{1dV}}.$$

• Interchanging the order of the iterated integrals does not change the answer. As a matter of fact, interchanging the order of integration can help simplify the computation.
• To compute the average value of a function over a general three-dimensional region, we use

• 交换累次积分的次序不会改变结果。事实上，交换积分次序还有助于简化计算。
• 为计算函数在一般三维区域上的平均值，我们使用

$$f_{\text{ave}} = \frac{1}{V(E)}{\iiint\limits_{E}{f\left( {x,y,z} \right)}}dV.$$

5.5 Triple Integrals in Cylindrical and Spherical Coordinates 5.5 柱坐标与球坐标下的三重积分

• To evaluate a triple integral in cylindrical coordinates, use the iterated integral

$${\int\limits_{\theta = \alpha}^{\theta = \beta}\ {\int\limits_{r = g_{1}{(\theta)}}^{r = g_{2}{(\theta)}}\ {\int\limits_{z = u_{1}{({r,\theta})}}^{z = u_{2}{({r,\theta})}}{f\left( {r,\theta,z} \right)}}}}r\ dz\ dr\ d\theta.$$

• To evaluate a triple integral in spherical coordinates, use the iterated integral

$${\int\limits_{\theta = \alpha}^{\theta = \beta}\ {\int\limits_{\rho = g_{1}{(\theta)}}^{\rho = g_{2}{(\theta)}}\ {\int\limits_{\varphi = u_{1}{({r,\theta})}}^{\varphi = u_{2}{({r,\theta})}}{f\left( {\rho,\theta,\varphi} \right)}}}}\rho^{2}\text{sin}\ \varphi\ d\varphi\ d\rho\ d\theta.$$

5.6 Calculating Centers of Mass and Moments of Inertia 5.6 计算质心与转动惯量

Finding the mass, center of mass, moments, and moments of inertia in double integrals:

• For a lamina $R$ with a density function $\rho\left( {x,y} \right)$ at any point $\left( {x,y} \right)$ in the plane, the mass is $m = {\iint\limits_{R}{\rho\left( {x,y} \right)dA}}.$
• The moments about the $x\text{-axis}$ and $y\text{-axis}$ are

• 对于平面上任意点 $\left( {x,y} \right)$ 处具有密度函数 $\rho\left( {x,y} \right)$ 的薄板 $R$，其质量为 $m = {\iint\limits_{R}{\rho\left( {x,y} \right)dA}}.$
• 关于 $x$ 轴与 $y$ 轴的矩为

$${M_{x} = {\iint\limits_{R}{y\rho\left( {x,y} \right)}}dA}\ \text{and}\ {M_{y} = {\iint\limits_{R}{x\rho\left( {x,y} \right)}}dA.}$$

• The center of mass is given by $\overset{\text{−}}{x} = \frac{M_{y}}{m},\overset{\text{−}}{y} = \frac{M_{x}}{m}.$
• The center of mass becomes the centroid of the plane when the density is constant.
• The moments of inertia about the $x - \text{axis,}$ $y - \text{axis,}$ and the origin are

• 质心由 $\overset{\text{−}}{x} = \frac{M_{y}}{m},\overset{\text{−}}{y} = \frac{M_{x}}{m}$ 给出。
• 当密度为常量时，质心成为该平面的形心。
• 关于 $x$ 轴、$y$ 轴以及原点的转动惯量为

$${I_{x} = {\iint\limits_{R}{y^{2}\rho\left( {x,y} \right)dA,}}\ \ I_{y} = {\iint\limits_{R}{x^{2}\rho\left( {x,y} \right)dA,}}}\ \text{and}\ {I_{0} = I_{x} + I_{y} = {\iint\limits_{R}{\left( {x^{2} + y^{2}} \right)\rho\left( {x,y} \right)dA.}}}$$

Finding the mass, center of mass, moments, and moments of inertia in triple integrals:

• For a solid object $Q$ with a density function $\rho\left( {x,y,z} \right)$ at any point $\left( {x,y,z} \right)$ in space, the mass is $m = {\iiint\limits_{Q}{\rho\left( {x,y,z} \right)dV}}.$
• The moments about the $xy\text{-plane,}$ the $xz\text{-plane,}$ and the $yz\text{-plane}$ are

• 对于空间中任意点 $\left( {x,y,z} \right)$ 处具有密度函数 $\rho\left( {x,y,z} \right)$ 的立体 $Q$，其质量为 $m = {\iiint\limits_{Q}{\rho\left( {x,y,z} \right)dV}}.$
• 关于 $xy$ 平面、$xz$ 平面与 $yz$ 平面的矩为

$${M_{xy} = {\iiint\limits_{Q}{z\rho\left( {x,y,z} \right)dV}},}\ {M_{xz} = {\iiint\limits_{Q}{y\rho\left( {x,y,z} \right)dV}},}\ {M_{yz} = {\iiint\limits_{Q}{x\rho\left( {x,y,z} \right)dV}}.}$$

• The center of mass is given by $\overset{\text{−}}{x} = \frac{M_{yz}}{m},\overset{\text{−}}{y} = \frac{M_{xz}}{m},\overset{\text{−}}{z} = \frac{M_{xy}}{m}.$
• The center of mass becomes the centroid of the solid when the density is constant.
• The moments of inertia about the $yz\text{-plane,}$ the $xz\text{-plane,}$ and the $xy\text{-plane}$ are

• 质心由 $\overset{\text{−}}{x} = \frac{M_{yz}}{m},\overset{\text{−}}{y} = \frac{M_{xz}}{m},\overset{\text{−}}{z} = \frac{M_{xy}}{m}$ 给出。
• 当密度为常量时，质心成为该立体的形心。
• 关于 $yz$ 平面、$xz$ 平面与 $xy$ 平面的转动惯量为

$$\begin{array}{l} \\ {I_{x} = {\iiint\limits_{Q}{\left( {y^{2} + z^{2}} \right)\rho\left( {x,y,z} \right)dV,}}I_{y} = {\iiint\limits_{Q}{\left( {x^{2} + z^{2}} \right)\rho\left( {x,y,z} \right)dV,}}} \\ {I_{z} = {\iiint\limits_{Q}{\left( {x^{2} + y^{2}} \right)\rho\left( {x,y,z} \right)dV.}}} \end{array}$$

5.7 Change of Variables in Multiple Integrals 5.7 多重积分中的变量代换

• A transformation $T$ is a function that transforms a region $G$ in one plane (space) into a region $R$ in another plane (space) by a change of variables.
• A transformation $T:G\rightarrow R$ defined as $T\left( {u,v} \right) = \left( {x,y} \right)$ $\left( {\text{or}\ T\left( {u,v,w} \right) = \left( {x,y,z} \right)} \right)$ is said to be a one-to-one transformation if no two points map to the same image point.
• If $f$ is continuous on $R,$ then ${\iint\limits_{R}{f\left( {x,y} \right)}}dA = {\iint\limits_{S}{f\left( {g\left( {u,v} \right),h\left( {u,v} \right)} \right)}}\left| \frac{\partial\left( {x,y} \right)}{\partial\left( {u,v} \right)} \right|du\ dv.$
• If $F$ is continuous on $R,$ then

• 变换 $T$ 是一个通过变量代换把一个平面（或空间）中的区域 $G$ 映为另一个平面（或空间）中的区域 $R$ 的函数。
• 若变换 $T:G\rightarrow R$ 定义为 $T\left( {u,v} \right) = \left( {x,y} \right)$ $\left( {\text{或}\ T\left( {u,v,w} \right) = \left( {x,y,z} \right)} \right)$，且没有两个不同的点映到同一个像点，则称它为一一对应变换。
• 若 $f$ 在 $R$ 上连续，则 ${\iint\limits_{R}{f\left( {x,y} \right)}}dA = {\iint\limits_{S}{f\left( {g\left( {u,v} \right),h\left( {u,v} \right)} \right)}}\left| \frac{\partial\left( {x,y} \right)}{\partial\left( {u,v} \right)} \right|du\ dv.$
• 若 $F$ 在 $R$ 上连续，则

$$\begin{array}{cl} {{\iiint\limits_{R}{F\left( {x,y,z} \right)}}dV} & {= {\iiint\limits_{G}{F\left( {g\left( {u,v,w} \right),h\left( {u,v,w} \right),k\left( {u,v,w} \right)} \right)}}\left| \frac{\partial\left( {x,y,z} \right)}{\partial\left( {u,v,w} \right)} \right|du\ dv\ dw} \\ & {= {\iiint\limits_{G}{H\left( {u,v,w} \right)}}\left| {J\left( {u,v,w} \right)} \right|du\ dv\ dw.} \end{array}$$

Review Exercises 复习题

*True or False?* Justify your answer with a proof or a counterexample.

416\.

${\int\limits_{a}^{b}\ {{\int\limits_{c}^{d}{f\left( {x,y} \right)dy\ dx}} =}}{\int\limits_{c}^{d}\ {\int\limits_{a}^{b}{f\left( {x,y} \right)dy\ dx}}}$

417.

Fubini's theorem can be extended to three dimensions, as long as $f$ is continuous in all variables.

418\.

The integral $\int\limits_{0}^{2\pi}\ {\int\limits_{0}^{1}\ {\int\limits_{r}^{1}{dz\ dr\ d\theta}}}$ represents the volume of a right cone.

419.

The Jacobian of the transformation $x = u^{2} - 2v,y = 3v - 2uv$ is given by $-4u^{2} + 6u + 4v.$

Evaluate the following integrals.

420\.

${\iint\limits_{R}\left( {5x^{3}y^{2} - y^{2}} \right)}dA,R = \left\{ {\left. \left( {x,y} \right) \right|0 \leq x \leq 2,1 \leq y \leq 4} \right\}$

421.

${\iint\limits_{D}\frac{y}{3x^{2} + 1}}dA,D = \left\{ {\left. \left( {x,y} \right) \right|0 \leq x \leq 1,\text{−}x \leq y \leq x} \right\}$

422\.

${\iint\limits_{D}{\text{sin}\left( {x^{2} + y^{2}} \right)}}dA$ where $D$ is a disk of radius $2$ centered at the origin

423.

$\int\limits_{0}^{1}\ {\int\limits_{y}^{1}{xye^{x^{2}}dx\ dy}}$

424\.

$\int\limits_{-1}^{1}\ {\int\limits_{0}^{z}\ {\int\limits_{0}^{x - z}{6dy\ dx\ dz}}}$

425.

${\iiint\limits_{R}{3y\ dV}},$ where $R = \left\{ {\left. \left( {x,y,z} \right) \right|0 \leq x \leq 1,0 \leq y \leq x,0 \leq z \leq \sqrt{9 - y^{2}}} \right\}$

426\.

$\int\limits_{0}^{2}\ {\int\limits_{0}^{2\pi}\ {\int\limits_{r}^{1}{r\ dz\ d\theta\ dr}}}$

427.

$\int\limits_{0}^{2\pi}\ {\int\limits_{0}^{\pi\text{/}2}\ {\int\limits_{1}^{3}{\rho^{2}\text{sin}(\varphi)d\rho\ d\varphi\ d\theta}}}$

428\.

$\int\limits_{0}^{1}\ {\int\limits_{\text{−}\sqrt{1 - x^{2}}}^{\sqrt{1 - x^{2}}}\ {{\int\limits_{\text{−}\sqrt{1 - x^{2} - y^{2}}}^{\sqrt{1 - x^{2} - y^{2}}}{dz}}\ dy\ dx}}$

For the following problems, find the specified area or volume.

429.

The area of region enclosed by one petal of $r = \text{cos}\left( {4\theta} \right).$

430\.

The volume of the solid that lies between the paraboloid $z = 2x^{2} + 2y^{2}$ and the plane $z = 8.$

431.

The volume of the solid bounded by the cylinder $x^{2} + y^{2} = 16$ and from $z = 1$ and $z + x = 2.$

432\.

The volume of the intersection between two spheres of radius 1, the top whose center is $(0,0,0.25)$ and the bottom, which is centered at $(0,0,0).$

For the following problems, find the center of mass of the region.

433.

$\rho(x,y) = xy$ on the circle with radius $1$ in the first quadrant only.

434\.

$\rho(x,y) = (y + 1)\sqrt{x}$ in the region bounded by $y = e^{x},$ $y = 0,$ and $x = 1.$

435.

$\rho(x,y,z) = z$ on the inverted cone with radius $2$ and height $2.$

436\.

The volume an ice cream cone that is given by the solid above $z = \sqrt{\left( {x^{2} + y^{2}} \right)}$ and below $z^{2} + x^{2} + y^{2} = z.$

The following problems examine Mount Holly in the state of Michigan. Mount Holly is a landfill that was converted into a ski resort. The shape of Mount Holly can be approximated by a right circular cone of height $1100$ ft and radius $6000$ ft.

437.

If the compacted trash used to build Mount Holly on average has a density $400{\ \text{lb/ft}}^{3},$ find the amount of work required to build the mountain.

438\.

In reality, it is very likely that the trash at the bottom of Mount Holly has become more compacted with all the weight of the above trash. Consider a density function with respect to height: the density at the top of the mountain is still density $400{\ \text{lb/ft}}^{3}$ and the density increases. Every $100$ feet deeper, the density doubles. What is the total weight of Mount Holly?

The following problems consider the temperature and density of Earth's layers.

439.

\[T\] The temperature of Earth's layers is exhibited in the table below. Use your calculator to fit a polynomial of degree $3$ to the temperature along the radius of the Earth. Then find the average temperature of Earth. (*Hint*: begin at $0$ in the inner core and increase outward toward the surface)

| Layer | Depth from center (km) | Temperature $\text{°}C$ |

|-------------------|------------------------|-------------------------|

| Rocky Crust | 0 to 40 | 0 |

| Upper Mantle | 40 to 150 | 870 |

| Mantle | 400 to 650 | 870 |

| Inner Mantel | 650 to 2700 | 870 |

| Molten Outer Core | 2890 to 5150 | 4300 |

| Inner Core | 5150 to 6378 | 7200 |

440.

\[T\] The density of Earth's layers is displayed in the table below. Using your calculator or a computer program, find the best-fit quadratic equation to the density. Using this equation, find the total mass of Earth.

| Layer | Depth from center (km) | Density (g/cm3) |

|--------------|------------------------|-----------------|

| Inner Core | $0$ | $12.95$ |

| Outer Core | $1228$ | $11.05$ |

| Mantle | $3488$ | $5.00$ |

| Upper Mantle | $6338$ | $3.90$ |

| Crust | $6378$ | $2.55$ |

The following problems concern the Theorem of Pappus (see Moments and Centers of Mass for a refresher), a method for calculating volume using centroids. Assuming a region $R,$ when you revolve around the $x\text{-axis}$ the volume is given by $V_{x} = 2\pi A\overset{–}{|y|},$ and when you revolve around the $\text{y-axis}$ the volume is given by $V_{y} = 2\pi A\left| \overset{–}{x} \right|,$ where $A$ is the area of $R.$ Consider the region bounded by $x^{2} + y^{2} = 1$ and above $y = x + 1.$

441.

Find the volume when you revolve the region around the $x\text{-axis.}$

442\.

Find the volume when you revolve the region around the $y\text{-axis.}$